Book 1. Foundations of Olympiad Geometry

Let the smaller angle be \(x\). Then the larger one is \(x+28^\circ\), and \(x+x+28^\circ=180^\circ\). Hence \(2x=152^\circ\), so \(x=76^\circ\). The other angle is \(104^\circ\).

Angles \(AOC\) and \(BOD\) are vertical, so \(\angle BOD=64^\circ\). Angle \(AOD\) is adjacent to \(AOC\), hence \(\angle AOD=180^\circ-64^\circ=116^\circ\). Angle \(BOC\) is vertical to \(AOD\), so \(\angle BOC=116^\circ\).

Let the angles be \(2x\), \(3x\), and \(4x\). Their sum is \(180^\circ\), hence \(9x=180^\circ\) and \(x=20^\circ\). The angles are \(40^\circ\), \(60^\circ\), and \(80^\circ\).

We have \(126^\circ=\angle A+\angle B\), so \(\angle B=126^\circ-49^\circ=77^\circ\). The interior angle at \(C\) is adjacent to the given exterior angle, hence \(\angle C=180^\circ-126^\circ=54^\circ\).

The alternate interior angle is \(117^\circ\). The same-side interior angle is \(180^\circ-117^\circ=63^\circ\).

Angles \(\angle CAB\) and \(\angle ABD\) are alternate interior angles for lines \(AC\), \(BD\) with transversal \(AB\). They are equal by the condition. By the parallelism criterion, \(AC\parallel BD\).

Let the same-side interior angles be \(\alpha\) and \(\beta\). Then \(\alpha+\beta=180^\circ\). Their bisectors make angles \(\frac{\alpha}{2}\) and \(\frac{\beta}{2}\) with the transversal. The sum of these angles is \(\frac{\alpha+\beta}{2}=90^\circ\). Therefore the angle between the bisectors is \(90^\circ\), so they are perpendicular.

Since \(l\parallel AC\), the two angles at line \(l\) are equal to angles \(A\) and \(C\) of the triangle, while the middle angle is \(\angle B\). Thus the triangle angles are in the ratio \(3:8:4\). Let them be \(3x\), \(8x\), and \(4x\). Then \(15x=180^\circ\), so \(x=12^\circ\). The angles are \(36^\circ\), \(96^\circ\), and \(48^\circ\).

Let two angles be \(x\) and \(y\), and the third be \(x+y\). The angle sum gives \(x+y+(x+y)=180^\circ\). Therefore \(2(x+y)=180^\circ\), so \(x+y=90^\circ\). The third angle is \(90^\circ\), hence the triangle is right-angled.

We have \(\angle A=180^\circ-42^\circ-74^\circ=64^\circ\). Since \(AD\) is the angle bisector, \(\angle BAD=32^\circ\). In triangle \(ABD\), \(\angle BDA=180^\circ-42^\circ-32^\circ=106^\circ\). Hence \(\angle ADC=180^\circ-106^\circ=74^\circ\).

Since \(AB=AC\), the base angles are equal: \(\angle B=\angle C=72^\circ\). Then \(\angle A=180^\circ-72^\circ-72^\circ=36^\circ\). The angle bisector \(BD\) gives \(\angle ABD=\angle DBC=36^\circ\). In triangle \(ABD\), the angles are \(36^\circ\), \(36^\circ\), \(108^\circ\). In triangle \(BCD\), \(\angle DBC=36^\circ\), \(\angle BCD=72^\circ\), hence \(\angle BDC=72^\circ\).

First, \(\angle A=180^\circ-54^\circ-62^\circ=64^\circ\). Since \(DE\parallel BC\), we have \(\angle ADE=\angle C=62^\circ\) and \(\angle AED=\angle B=54^\circ\). Therefore the angles of triangle \(ADE\) are \(64^\circ\), \(62^\circ\), and \(54^\circ\).

Let \(\angle B=\beta\), \(\angle C=\gamma\), and assume \(\beta>\gamma\). Then \(\angle A=180^\circ-\beta-\gamma\), so \(\angle BAL=\frac{180^\circ-\beta-\gamma}{2}\). Since \(AH\perp BC\), the angle between \(AB\) and \(AH\) is \(90^\circ-\beta\). Thus the angle between \(AH\) and \(AL\) equals \(\frac{180^\circ-\beta-\gamma}{2}-(90^\circ-\beta)=\frac{\beta-\gamma}{2}\). If \(\gamma>\beta\), the same argument gives \(\frac{\gamma-\beta}{2}\).

In triangle \(BIC\), \(\angle IBC=\frac{B}{2}\) and \(\angle ICB=\frac{C}{2}\). Hence \(\angle BIC=180^\circ-\frac{B+C}{2}\). Since \(B+C=180^\circ-A\), we get \(\angle BIC=180^\circ-\frac{180^\circ-A}{2}=90^\circ+\frac{A}{2}\). Thus \(124^\circ=90^\circ+\frac{A}{2}\), so \(A=68^\circ\).

Let the exterior bisector make an angle \(x\) with side \(AB\). Since it is parallel to \(BC\), this angle equals \(\angle B\). Since it bisects the exterior angle, the second angle between it and line \(AC\) is also \(x\). By parallelism, it equals \(\angle C\). Therefore \(\angle B=\angle C\). The triangle is isosceles, so \(AB=AC\).

In triangle \(ABC\), \(\angle C=180^\circ-60^\circ-50^\circ=70^\circ\). Since \(D\) lies on \(AB\), angle \(CBD\) equals \(\angle CBA=50^\circ\). In triangle \(BCD\), sides \(CD\) and \(BD\) are equal, so the base angles at \(BC\) are equal: \(\angle BCD=\angle CBD=50^\circ\). Hence \(\angle ACD=\angle ACB-\angle BCD=70^\circ-50^\circ=20^\circ\).

Since \(AB=AC\), the base angles are equal: \(\angle B=\angle C=\frac{180^\circ-36^\circ}{2}=72^\circ\). In triangle \(BCD\), sides \(BD\) and \(BC\) are equal, so \(\angle BDC=\angle BCD\). Since \(D\) lies on \(AC\), \(\angle BCD=72^\circ\), hence \(\angle BDC=72^\circ\). Therefore \(\angle DBC=180^\circ-72^\circ-72^\circ=36^\circ\). Thus \(\angle ABD=\angle ABC-\angle DBC=72^\circ-36^\circ=36^\circ\).

Angles \(\angle BAC\) and \(\angle DCA\) are alternate interior angles for lines \(AB\) and \(CD\) with transversal \(AC\). They are equal, so \(AB\parallel CD\). Similarly, angles \(\angle BCA\) and \(\angle DAC\) are alternate interior angles for lines \(BC\) and \(AD\) with transversal \(AC\). They are equal, hence \(BC\parallel AD\).

Since \(DE\parallel AB\), the angle between \(DE\) and \(AD\) equals \(\angle BAD\). Since \(DF\parallel AC\), the angle between \(AD\) and \(DF\) equals \(\angle DAC\). By the condition, \(\angle BAD=\angle DAC\). Hence \(AD\) divides \(\angle EDF\) into two equal angles, so it is its angle bisector.

Since \(DE\parallel AB\), the angle between \(DE\) and \(AD\) equals \(\angle BAD\). Since \(DF\parallel AC\), the angle between \(AD\) and \(DF\) equals \(\angle DAC\). By the condition, \(AD\) bisects \(\angle EDF\), so these two angles at \(D\) are equal. Therefore \(\angle BAD=\angle DAC\), meaning that \(AD\) bisects angle \(A\).

In triangles \(ABC\) and \(DEF\), two sides and the included angle are respectively equal. By SAS, the triangles are congruent. Therefore the corresponding third sides are equal: \(BC=EF\).

A side and its two adjacent angles are respectively equal in the two triangles. Hence the triangles are congruent by ASA. Therefore the corresponding sides \(AC\) and \(A_1C_1\) are equal.

Since \(AB=AC\), we have \(\angle B=\angle C\). Their sum is \(180^\circ-52^\circ=128^\circ\). Therefore \(\angle B=\angle C=64^\circ\).

We have \(AB=AC\) by the condition, \(BM=CM\) because \(M\) is the midpoint of \(BC\), and \(AM\) is common. Hence \(\triangle ABM=\triangle ACM\) by SSS.

Angles \(AOB\) and \(COD\) are vertical, so they are equal. Also \(AO=CO\) and \(BO=DO\). By SAS, \(\triangle AOB=\triangle COD\). Therefore the corresponding sides \(AB\) and \(CD\) are equal.

By SSS, triangles \(ABM\) and \(ACM\) are congruent: \(AB=AC\), \(BM=CM\), and \(AM\) is common. Therefore \(\angle BAM=\angle MAC\), so \(AM\) is an angle bisector. Also \(\angle AMB=\angle AMC\). These angles are adjacent, so both are \(90^\circ\). Hence \(AM\perp BC\), so \(AM\) is an altitude.

Since \(M\) is the midpoint of \(BC\), we have \(BM=CM\). Side \(AM\) is common, and angles \(AMB\) and \(AMC\) are right angles. By SAS, triangles \(ABM\) and \(ACM\) are congruent. Therefore \(AB=AC\).

In triangles \(ABM\) and \(ACM\), we have \(BM=CM\), common side \(AM\), and right angles \(AMB\) and \(AMC\). By SAS, the triangles are congruent. Therefore \(AB=AC\).

Since \(AM\) is a median, \(BM=CM\). By the condition, \(AB+BM+AM=AC+CM+AM\). Cancelling equal terms \(BM=CM\) and common side \(AM\), we get \(AB=AC\).

In triangles \(ABN\) and \(ACM\), we have \(AB=AC\), \(AN=AM\), and the included angle in both triangles is angle \(A\). By SAS, the triangles are congruent. Therefore the corresponding sides \(BN\) and \(CM\) are equal.

From the congruence of the original triangles, \(AB=A_1B_1\), \(BC=B_1C_1\), and \(\angle ABC=\angle A_1B_1C_1\). Since \(M\) and \(M_1\) are midpoints, \(BM=\frac{1}{2}BC\) and \(B_1M_1=\frac{1}{2}B_1C_1\), so \(BM=B_1M_1\). By SAS, \(\triangle ABM=\triangle A_1B_1M_1\). Therefore \(AM=A_1M_1\).

Consider triangles \(ABC\) and \(CDA\). They have \(AB=CD\), \(BC=AD\), and common side \(AC\). By SSS, the triangles are congruent. Therefore the corresponding angles \(ABC\) and \(CDA\) are equal.

We have \(AM=MP\) and \(BM=CM\). Angles \(AMB\) and \(PMC\) are vertical, so they are equal. By SAS, \(\triangle AMB=\triangle PMC\), hence \(AB=PC\). Similarly, angles \(AMC\) and \(PMB\) are vertical, with \(AM=MP\) and \(CM=BM\). Thus \(\triangle AMC=\triangle PMB\), so \(AC=PB\).

Since \(AB=AC\) and \(BD=CE\), we get \(AD=AE\). Hence \(\triangle ADE\) is isosceles. Triangle \(ABC\) is also isosceles. The two triangles share the angle at \(A\), so their base angles are equal: \(\angle ADE=\angle ABC\). These are corresponding angles for lines \(DE\) and \(BC\), therefore \(DE\parallel BC\).

Angles \(AOB\) and \(COD\) are vertical, so they are equal. Together with \(AO=OC\) and \(BO=OD\), this gives \(\triangle AOB=\triangle COD\) by SAS. Hence \(\angle ABO=\angle CDO\), and equal alternate interior angles imply \(AB\parallel CD\). Similarly, from congruent triangles \(AOD\) and \(COB\), we get \(AD\parallel BC\).

Triangles \(ABD\) and \(ACD\) are right triangles, have common side \(AD\), and \(\angle BAD=\angle DAC\) because \(AD\) is an angle bisector. By ASA, the triangles are congruent. Therefore \(AB=AC\).

Since the points lie on the same rays, \(AD=AB+BD\) and \(AE=AC+CE\). From \(AB=AC\) and \(BD=CE\), we get \(AD=AE\). Triangles \(ABC\) and \(ADE\) are isosceles with the same angle at \(A\), so their base angles are equal. Hence \(\angle ADE=\angle ABC\), which gives \(DE\parallel BC\).

In triangles \(ABC\) and \(ADC\), we have \(AB=AD\), \(CB=CD\), and common side \(AC\). By SSS, the triangles are congruent. Therefore \(\angle BAC=\angle CAD\) and \(\angle BCA=\angle ACD\). Hence \(AC\) bisects angles \(A\) and \(C\).

Triangles \(ABH\) and \(ACK\) are right triangles. They have \(BH=CK\) by the condition and \(\angle BAH=\angle CAK\), because both are equal to angle \(A\) of the original triangle. Therefore the third angles are also equal: \(\angle ABH=\angle ACK\). Sides \(BH\) and \(CK\) lie between equal angles, so by ASA triangles \(ABH\) and \(ACK\) are congruent. Therefore their hypotenuses are equal: \(AB=AC\).

Since \(M\) is the midpoint of \(AB\), \(AM=BM\). Also \(MC=MD\). Angles \(AMC\) and \(BMD\) are vertical, so they are equal. By SAS, \(\triangle AMC=\triangle BMD\), hence \(AC=BD\). Similarly, \(\angle AMD=\angle BMC\), and by SAS \(\triangle AMD=\triangle BMC\), so \(AD=BC\).

Consider triangles \(ABC\) and \(ADC\). They have \(AB=AD\), \(CB=CD\), and common side \(AC\), so they are congruent by SSS. Therefore \(\angle BAC=\angle CAD\). Since \(O\) lies on \(AC\), we have \(\angle BAO=\angle OAD\). In triangles \(ABO\) and \(ADO\), \(AB=AD\), \(AO\) is common, and \(\angle BAO=\angle OAD\). By SAS, these triangles are congruent. Hence \(BO=DO\) and \(\angle AOB=\angle AOD\). Angles \(AOB\) and \(AOD\) are adjacent, so each is \(90^\circ\). Therefore \(AC\perp BD\).

Since \(P\) lies on \(AB\) and \(Q\) on \(AC\), we have \(AB=AP+PB\) and \(AC=AQ+QC\). From \(AP=AQ\) and \(BP=CQ\), it follows that \(AB=AC\). Hence triangle \(ABC\) is isosceles, and median \(AM\), drawn to base \(BC\), is the angle bisector of angle \(A\). Now consider triangles \(APM\) and \(AQM\). They have \(AP=AQ\), common side \(AM\), and \(\angle PAM=\angle MAQ\). By SAS, the triangles are congruent, so \(PM=QM\).

E. Full solution.

The segment of length \(a_1\) belonged to one of the two original triangles. Let the other two sides of that triangle have lengths \(u\) and \(v\). By the triangle inequality, \(a_1

Among the five segments different from \(a_1\), the two longest have lengths \(a_2\) and \(a_3\). Hence \(a_2+a_3\ge u+v\). Therefore \(a_1

Since \(a_1\ge a_2\ge a_3\), the other two triangle inequalities are immediate: \(a_2

The statement for the three shortest segments is false. For example, take two congruent triangles with side lengths \(1,3,3\). Then the three shortest lengths are \(1,1,3\), which cannot form a triangle.

E. Full solution.

We prove the statement by induction on the number of sides. For a triangle it is immediate, because the triangle itself is isosceles.

Assume the statement is known for polygons with fewer sides. Consider the convex polygon \(P\) and one of the drawn diagonals \(d\). It divides \(P\) into two smaller convex polygons \(P_1\) and \(P_2\). Each of them is also triangulated into isosceles triangles.

By the induction hypothesis, \(P_1\) has two equal sides. If both are sides of the original polygon, we are done. If one of them is the diagonal \(d\), then \(P_1\) has a side of the original polygon equal to \(d\).

The same argument for \(P_2\) either immediately gives two equal sides of the original polygon, or gives a side of the original polygon equal to \(d\). In the last case the two found sides lie on opposite sides of the diagonal and are both equal to \(d\), hence they are equal to each other.

E. Full solution.

Let \(U\) be the midpoint of side \(PS\). Since \(PX=SX\), in isosceles triangle \(PXS\) the median \(XU\) to the base is also an altitude. Hence \(XU\perp PS\). Since \(PS\parallel QR\), we get \(XU\perp QT\).

Because \(U\) and \(T\) are midpoints of opposite sides of the parallelogram, quadrilaterals \(PQTU\) and \(QTSU\) are parallelograms. Thus \(UT\parallel PQ\) and \(QU\parallel ST\).

By the condition \(QX\perp PQ\), and since \(UT\parallel PQ\), we have \(QX\perp UT\). Therefore in triangle \(QUT\), the lines \(XU\) and \(XQ\) are two altitudes. Hence \(X\) is the orthocenter of triangle \(QUT\).

The third altitude of this triangle also passes through \(X\), so \(XT\perp QU\). But \(QU\parallel ST\), hence \(XT\perp ST\), as required.

E. Full solution.

Take an isosceles triangle that is not equilateral, and glue two copies along a lateral side so that a convex kite is formed. Its adjacent sides are equal in pairs, but opposite sides need not be parallel.

Choose the original isosceles triangle so that its vertex angle is not equal to its base angle. Then in the resulting kite no pair of opposite sides is parallel: otherwise equality of corresponding angles with a transversal would force the vertex angle to equal a base angle.

Now draw the altitude to the base in each of the two isosceles triangles. In an isosceles triangle this altitude splits it into two congruent right triangles. The two larger isosceles triangles were congruent, so all four small right triangles are pairwise congruent.

Thus the same kite is assembled from four congruent triangles and has no parallel sides. Therefore the answer is yes.

E. Full solution.

We first show that ten unit segments are impossible. If all ten segments had length \(1\), then triangles \(OAB\), \(OBC\), \(OCD\), \(ODE\), and \(OEA\) would all be equilateral.

Hence each of the angles \(\angle AOB\), \(\angle BOC\), \(\angle COD\), \(\angle DOE\), and \(\angle EOA\) would be \(60^\circ\). Their sum around point \(O\) must be \(360^\circ\), but it would be \(5\cdot60^\circ=300^\circ\), a contradiction. Thus at most nine are possible.

Now construct an example with nine. Take a regular hexagon of side \(1\) with center \(O\). Choose five consecutive vertices as \(A,B,C,D,E\). Then \(O\) lies inside the pentagon.

In this example all five segments \(OA,OB,OC,OD,OE\) are equal to \(1\), and four sides \(AB,BC,CD,DE\) are also equal to \(1\). The remaining side \(EA\) is not \(1\). Therefore the maximum is exactly nine.

E. Full solution.

Let \(H\) be the foot of the perpendicular from \(D\) to \(BC\). Since \(\angle BCD=60^\circ\), right triangle \(DCH\) has a \(60^\circ\) angle at \(C\). Therefore \(CH=\frac{CD}{2}=AB\).

Denote \(\angle CBM=\alpha\). Since \(BD\) is the exterior angle bisector at \(B\), the angle between \(BA\) and \(BM\) is \(180^\circ-\alpha\).

In right triangle \(BHD\), point \(M\) is the midpoint of hypotenuse \(BD\). Hence \(MH=MB\). Therefore \(\angle MHB=\angle HBM=\alpha\), and so \(\angle MHC=180^\circ-\alpha\).

Now compare triangles \(ABM\) and \(CHM\). We have \(AB=CH\), \(BM=HM\), and the included angles are equal: \(\angle ABM=180^\circ-\alpha=\angle CHM\). Thus the triangles are congruent by SAS.

From the congruence, \(AM=CM\).

E. Full solution.

Reflect \(H\) across \(AB\), and call the image \(C'\). Then \(E\) is the midpoint of \(HC'\), and \(BC'=BH\), because reflection preserves distances and point \(B\) lies on the axis of reflection.

Similarly, reflect \(H\) across \(AC\), and call the image \(B'\). Then \(D\) is the midpoint of \(HB'\), and \(CB'=CH\).

In triangle \(ADE\), line \(DF\) is perpendicular to \(AE\), and \(AE\subset AB\), so \(DF\parallel CE\). Similarly, \(EF\parallel BD\). Hence \(H E F D\) is a parallelogram.

In triangle \(HC'B'\), points \(E\) and \(D\) are the midpoints of sides \(HC'\) and \(HB'\). Therefore the parallelogram \(H E F D\) shows that \(F\) is the midpoint of side \(C'B'\).

Points \(M\) and \(F\) are the midpoints of \(BC\) and \(B'C'\). Thus \(2FM\le BC'+CB'\); this is the triangle inequality applied to the sum of the two vectors from \(B\) to \(C'\) and from \(C\) to \(B'\).

But \(BC'=BH\) and \(CB'=CH\). Therefore \(2FM\le BH+CH\), as required.

By AA, \(\triangle ABC\sim\triangle DEF\). Side \(AB\) corresponds to \(DE\), so the ratio is \(\frac{15}{5}=3\). Side \(BC\) corresponds to \(EF\), hence \(EF=7\cdot3=21\).

Since \(AB:A_1B_1=3:4\), we also have \(AC:A_1C_1=3:4\). Therefore \(A_1C_1=12\cdot\frac{4}{3}=16\).

Segment \(MN\) is a midline of triangle \(ABC\). Therefore \(MN=\frac{1}{2}BC=9\).

If the sides are in the ratio \(2:3\), the areas are in the ratio \(4:9\). Therefore the area of the larger triangle is \(20\cdot\frac{9}{4}=45\).

Since \(DE\parallel BC\), \(\triangle ADE\sim\triangle ABC\). Hence \(\frac{DE}{BC}=\frac{AD}{AB}=\frac{3}{5}\). Therefore \(DE=25\cdot\frac{3}{5}=15\).

We have \(AB=AD+DB=10\), so \(\frac{AD}{AB}=\frac{4}{10}=\frac{2}{5}\). From \(\triangle ADE\sim\triangle ABC\), we get \(\frac{AE}{AC}=\frac{2}{5}\). Thus \(AC=5\cdot\frac{5}{2}=\frac{25}{2}\). Therefore \(EC=AC-AE=\frac{25}{2}-5=\frac{15}{2}\).

We have \(\frac{AB}{DE}=\frac{6}{9}=\frac{2}{3}\) and \(\frac{AC}{DF}=\frac{10}{15}=\frac{2}{3}\). The included angles are equal. Hence the triangles are similar by SAS.

We have \(\frac{6}{9}=\frac{8}{12}=\frac{10}{15}=\frac{2}{3}\). All three sides are proportional, so the triangles are similar by SSS.

Since \(M\) and \(N\) are midpoints, \(\frac{AM}{AB}=\frac{AN}{AC}=\frac{1}{2}\). Angle \(A\) is common. By SAS similarity, triangles \(AMN\) and \(ABC\) are similar with ratio \(\frac{1}{2}\). Therefore \(MN\) corresponds to \(BC\), so \(MN\parallel BC\) and \(MN=\frac{1}{2}BC\).

From the similarity of triangles \(ACH\) and \(CBH\), \(\frac{CH}{AH}=\frac{HB}{CH}\). Therefore \(CH^2=AH\cdot HB=3\cdot12=36\), so \(CH=6\).

Let the height of the tower be \(h\). By similarity of right triangles, \(\frac{h}{18}=\frac{2.4}{3}\). Hence \(h=18\cdot0.8=14.4\) m.

The areas of similar triangles are in the ratio of the squares of corresponding sides. Therefore the sides are in the ratio \(\sqrt{25}:\sqrt{49}=5:7\).

Since \(AD:DB=2:1\), we have \(AD:AB=2:3\). From \(DE\parallel BC\), \(\triangle ADE\sim\triangle ABC\) with ratio \(\frac{2}{3}\). Areas are in the ratio \(\frac{4}{9}\). Therefore \(S_{ADE}=54\cdot\frac{4}{9}=24\).

Triangles \(ABC\) and \(ACH\) are right triangles: \(\angle C=90^\circ\), \(\angle AHC=90^\circ\). Also, the angle at \(A\) is common. Therefore \(\triangle ABC\sim\triangle ACH\). The side correspondence gives \(\frac{AB}{AC}=\frac{AC}{AH}\). Hence \(AC^2=AB\cdot AH\).

From \(B_1C_1\parallel B_2C_2\), \(\triangle AB_1C_1\sim\triangle AB_2C_2\). Thus \(\frac{AB_1}{AB_2}=\frac{AC_1}{AC_2}\). We get \(\frac{6}{10}=\frac{AC_1}{15}\), so \(AC_1=9\).

Let \(P\) and \(Q\) be the midpoints of \(BG\) and \(CG\). Then \(PQ\parallel BC\) and \(PQ=\frac{1}{2}BC\). Since \(M\) and \(N\) are the midpoints of \(AC\) and \(AB\), segment \(MN\) is also parallel to \(BC\) and equals \(\frac{1}{2}BC\). Hence \(PQ\parallel MN\) and \(PQ=MN\), so \(MNPQ\) is a parallelogram. Its diagonals \(MP\) and \(NQ\) meet at \(G\) and bisect each other. Therefore \(GM=GP\), and since \(P\) is the midpoint of \(BG\), \(BP=PG\). Thus \(BG=2GM\). Similarly, \(CG=2GN\).

Since \(PE\parallel AB\), triangles \(CPE\) and \(CAB\) are similar. From \(AP:PC=2:3\), we get \(CP:CA=3:5\). Therefore \(CE:CB=3:5\). Since \(EB=CB-CE\), \(CE:EB=3:(5-3)=3:2\).

By similarity of triangles \(ABC\) and \(ACH\), \(AC^2=AB\cdot AH\). Therefore \(AC^2=25\cdot9=225\), so \(AC=15\).

Let the side of the square be \(x\). The top side of the square is parallel to the base of the triangle and lies at distance \(x\) from the base. The small triangle above the square is similar to the original triangle; its altitude is \(h-x\), so the length of the top side of the square is \(a\cdot\frac{h-x}{h}\). But this length equals \(x\). Thus \(x=a\cdot\frac{h-x}{h}\), or \(xh=ah-ax\). Hence \(x(a+h)=ah\), so \(x=\frac{ah}{a+h}\).

Let \(K\) and \(L\) be the midpoints of sides \(BC\) and \(AB\); the medians are \(AK\) and \(CL\). It is enough to prove the statement in a convenient affine model, because parallelism and ratios on one line are preserved. Take coordinates \(A=(0,0)\), \(C=(1,0)\), \(B=(0,1)\). Then \(K=\left(\frac{1}{2},\frac{1}{2}\right)\), \(L=\left(0,\frac{1}{2}\right)\). Let \(P=(t,0)\).

The line through \(P\) parallel to \(AK\) has direction \((1,1)\) and meets \(BC\) at \(E=\left(\frac{1+t}{2},\frac{1-t}{2}\right)\). The line through \(P\) parallel to \(CL\) has direction \((-1,\frac{1}{2})\) and meets \(AB\) at \(F=\left(0,\frac{t}{2}\right)\). Parametrize \(EF\) as \(Z(\lambda)=E+\lambda(F-E)\).

Median \(AK\) has equation \(y=x\). Substituting \(Z(\lambda)\) gives \(\lambda=\frac{1}{3}\). Median \(CL\) has equation \(y=\frac{1-x}{2}\). Substitution gives \(\lambda=\frac{2}{3}\). Thus the intersection points of the medians with \(EF\) divide the segment at \(\frac{1}{3}\) and \(\frac{2}{3}\), so the three parts are equal.

E. Full solution.

By the tangent-chord theorem, \(\angle PBA=\angle BCA\), and the angle at \(P\) is common to triangles \(PBA\) and \(PCB\). Hence these triangles are similar, so \(\frac{AB}{BC}=\frac{PB}{PC}\).

Similarly, triangles \(PDA\) and \(PCD\) are similar, giving \(\frac{AD}{DC}=\frac{PD}{PC}\). But tangents from the same point are equal: \(PB=PD\). Therefore \(\frac{AB}{BC}=\frac{AD}{DC}\).

Now suppose the line through \(X\), parallel to \(BD\), meets sides \(AB\) and \(AD\). Then by similarity, \(\frac{AY}{AB}=\frac{AZ}{AD}=\frac{AX}{AC}\), so \(Y\) and \(Z\) divide the first links of the broken paths in the same ratio.

If instead the line meets sides \(BC\) and \(DC\), then similarly \(\frac{CY}{CB}=\frac{CZ}{CD}=\frac{CX}{CA}\), so the remaining parts of both broken paths have the same relative size.

Since the two full broken paths have the same ratio between corresponding links, in both cases the point on one broken path and the point on the other represent the same fraction of the path from \(A\) to \(C\).

Draw diagonal \(AC\). Since \(AB\parallel CD\), we have \(\angle BAC=\angle ACD\). Since \(AD\parallel BC\), we have \(\angle DAC=\angle ACB\). Adding the equal parts gives \(\angle A=\angle C\). Similarly, drawing diagonal \(BD\) gives \(\angle B=\angle D\).

Draw \(AC\). From \(AB\parallel CD\), \(\angle BAC=\angle ACD\); from \(AD\parallel BC\), \(\angle BCA=\angle CAD\). Side \(AC\) is common. Thus \(\triangle ABC\cong\triangle CDA\) by a side and two adjacent angles. Therefore \(AB=CD\) and \(BC=AD\).

Let \(ABCD\) be a rectangle. Consider triangles \(ABC\) and \(DCB\). We have \(AB=CD\) because a rectangle is a parallelogram, side \(BC\) is common, and angles \(ABC\) and \(DCB\) are right. Thus the triangles are congruent, so \(AC=BD\).

In a rhombus all sides are equal, so \(AB=AD\) and \(BC=CD\). Side \(AC\) is common. Hence \(\triangle ABC\cong\triangle ADC\) by SSS. Therefore \(\angle BAC=\angle CAD\) and \(\angle BCA=\angle ACD\). Diagonal \(AC\) bisects angles \(A\) and \(C\).

By the trapezoid midline theorem, \(MN=\frac{AD+BC}{2}=\frac{17+9}{2}=13\).

Draw \(AC\). Since \(AB\parallel CD\), \(\angle BAC=\angle ACD\). Also \(AB=CD\), and \(AC\) is common. Thus \(\triangle BAC\cong\triangle DCA\) by SAS. Therefore \(\angle BCA=\angle CAD\), so \(BC\parallel AD\). Together with \(AB\parallel CD\), this gives a parallelogram.

Triangles \(AOB\) and \(COD\) are congruent by SAS: \(AO=OC\), \(BO=OD\), and angles \(AOB\) and \(COD\) are vertical. Hence \(\angle ABO=\angle CDO\), so \(AB\parallel CD\). Similarly, triangles \(AOD\) and \(COB\) are congruent, so \(AD\parallel BC\). Both pairs of opposite sides are parallel, hence \(ABCD\) is a parallelogram.

Let \(\angle A=90^\circ\) in parallelogram \(ABCD\). Since \(AD\parallel BC\), angles \(A\) and \(B\) are same-side interior angles, so \(\angle A+\angle B=180^\circ\). Hence \(\angle B=90^\circ\). Opposite angles of a parallelogram are equal, so \(\angle C=90^\circ\) and \(\angle D=90^\circ\). All angles are right, so it is a rectangle.

In a parallelogram, \(AB=CD\) and \(BC=AD\). By condition \(AB=BC\). Hence \(AB=BC=CD=AD\). All sides are equal, so \(ABCD\) is a rhombus.

Let \(MN\) be the midline. Then \(14=\frac{AD+8}{2}\). Hence \(AD+8=28\), so \(AD=20\).

Since \(AB=CD\), the trapezoid is isosceles, so \(\angle DAB=\angle CDA\). From \(AD\parallel BC\), we get \(\angle DAB+\angle ABC=180^\circ\). Replacing \(\angle DAB\) by equal angle \(\angle CDA\), we obtain \(\angle ABC+\angle CDA=180^\circ\). The sum of opposite angles is \(180^\circ\), hence quadrilateral \(ABCD\) is cyclic.

We have \(\angle A+\angle C=74^\circ+106^\circ=180^\circ\). In a convex quadrilateral, if the sum of opposite angles is \(180^\circ\), then its vertices lie on one circle. Therefore \(A,B,C,D\) lie on one circle.

In triangle \(ABC\), segment \(MN\) joins the midpoints of two sides, so \(MN\parallel AC\) and \(MN=\frac{1}{2}AC\). In triangle \(CDA\), segment \(PQ\) also joins the midpoints of two sides, so \(PQ\parallel AC\) and \(PQ=\frac{1}{2}AC\). Therefore \(MN\parallel PQ\) and \(MN=PQ\). If one pair of opposite sides in a quadrilateral is both equal and parallel, then it is a parallelogram. Hence \(MNPQ\) is a parallelogram.

By the midline theorem, \(MN=PQ=\frac{1}{2}AC=\frac{11}{2}\), and \(NP=QM=\frac{1}{2}BD=\frac{15}{2}\). Thus the perimeter is \(2\cdot\frac{11}{2}+2\cdot\frac{15}{2}=11+15=26\).

Triangles \(AOD\) and \(COB\) are similar: \(\angle AOD=\angle COB\) as vertical angles, and \(\angle ADO=\angle CBO\) because \(AD\parallel BC\). Therefore \(\frac{AO}{OC}=\frac{DO}{OB}=\frac{AD}{BC}=\frac{21}{14}=\frac{3}{2}\). Hence \(AO:OC=DO:OB=3:2\).

Since \(MN\parallel BC\), in triangle \(ABC\) the line through midpoint \(M\) of side \(AB\), parallel to \(BC\), meets \(AC\) at its midpoint. Hence \(P\) is the midpoint of \(AC\). Then \(MP\) is a midline of triangle \(ABC\), so \(MP=\frac{1}{2}BC\). Also \(PN\) is a midline of triangle \(ACD\), so \(PN=\frac{1}{2}AD\). Therefore \(MP:PN=BC:AD\).

By condition \(\angle BAC=\angle CAD\). Since \(AD\parallel BC\), \(\angle CAD=\angle ACB\). Therefore \(\angle BAC=\angle ACB\), so in triangle \(ABC\), sides \(AB\) and \(BC\) are equal. In a parallelogram, equality of adjacent sides means that all sides are equal. Hence \(ABCD\) is a rhombus.

From \(AO=OC\) and \(BO=OD\), the diagonals bisect each other. Hence \(ABCD\) is a parallelogram. A parallelogram with equal diagonals has all angles right: this follows by comparing triangles \(ABC\) and \(DCB\). Therefore \(ABCD\) is a rectangle.

Angle \(\angle CDE\) is formed by rays \(DC\) and \(DE\), and angle \(\angle CBE\) by rays \(BC\) and \(BE\). Both angles stand on the same segment \(CE\). If two points \(B\) and \(D\) see segment \(CE\) under equal angles, then they lie on one circle with \(C\) and \(E\). Therefore \(B,C,D,E\) lie on one circle.

In triangle \(ABC\), segment \(EP\parallel AB\), where \(E\) lies on \(BC\) and \(P\) lies on \(AC\). Therefore \(\triangle CPE\sim\triangle CAB\). From \(BE:EC=1:2\), we have \(CE:CB=2:3\). Hence \(CP:CA=2:3\). Thus \(AP:PC=(CA-CP):CP=1:2\).

Since \(M\) is the midpoint of \(AB\) and \(MN\parallel BC\), in triangle \(ABC\), point \(P\) is the midpoint of \(AC\), and \(MP=\frac{1}{2}BC\). Similarly, in triangle \(ABD\), the line through midpoint \(M\), parallel to \(AD\), meets \(BD\) at its midpoint \(Q\), and \(MQ=\frac{1}{2}AD\). Since \(AD>BC\), point \(Q\) lies farther from \(M\) than \(P\), so \(PQ=MQ-MP=\frac{AD}{2}-\frac{BC}{2}=\frac{AD-BC}{2}\).

Let \(AD=a\), \(BC=b\). From the similarity of triangles \(AOD\) and \(COB\), we get \(DO:OB=a:b\), so \(OB:DB=b:(a+b)\) and \(DO:DB=a:(a+b)\). In triangle \(BDA\), segment \(OX\parallel AD\), hence \(\frac{OX}{AD}=\frac{OB}{DB}=\frac{b}{a+b}\). Thus \(OX=\frac{ab}{a+b}\). In triangle \(DBC\), segment \(OY\parallel BC\), hence \(\frac{OY}{BC}=\frac{DO}{DB}=\frac{a}{a+b}\). Thus \(OY=\frac{ab}{a+b}\). Therefore \(OX=OY\). For \(a=12\), \(b=6\), we get \(OX=OY=\frac{12\cdot6}{18}=4\), so \(XY=8\).

In a parallelogram, \(\angle A=\angle C\). Since the quadrilateral is cyclic, opposite angles are supplementary, so \(\angle A+\angle C=180^\circ\). But these angles are equal, hence each is \(90^\circ\). Similarly, the remaining angles are also right. Therefore \(ABCD\) is a rectangle.

In an isosceles trapezoid, base angles are equal, so triangles \(DAB\) and \(ADC\) are congruent by two sides and the included angle: \(AD\) is common, \(AB=CD\), and the base angles at \(AD\) are equal. Therefore \(BD=AC\). Since the diagonals of a trapezoid are divided in the ratio of the bases, \(AO:OC=DO:OB=AD:BC=18:10=9:5\). The diagonals \(AC\) and \(BD\) are equal and divided in the same ratio \(9:5\), so their larger parts are equal: \(AO=DO\). Answer: \(AO:OC=9:5\).

Angle \(\angle AOB\) is central, and \(\angle ACB\) is inscribed. They stand on the same arc \(AB\). Therefore \(\angle ACB=\frac{1}{2}\cdot96^\circ=48^\circ\).

Angles \(\angle ABC\) and \(\angle ADC\) are inscribed and stand on the same chord \(AC\). With the given position, they look at the same arc \(AC\), so they are equal.

The central angle corresponding to diameter \(AB\) is \(180^\circ\). The inscribed angle \(\angle ACB\) standing on the same arc is \(90^\circ\). Therefore \(AC\perp BC\).

We have \(OA=OB=OC=OD\), since these are radii of one circle. By condition \(AB=CD\). Hence \(\triangle AOB\cong\triangle COD\) by SSS, so \(\angle AOB=\angle COD\).

By the tangent theorem, the radius drawn to the point of tangency is perpendicular to the tangent. Since \(A\) is the point of tangency, \(OA\perp l\).

Since \(ABCD\) is cyclic, \(\angle A+\angle C=180^\circ\). Therefore \(\angle C=180^\circ-68^\circ=112^\circ\).

Both angles are inscribed and stand on chord \(AC\). With the given position, they are equal. Hence \(\angle ADC=41^\circ\).

The angle between the tangent at \(A\) and chord \(AC\) equals the inscribed angle standing on chord \(AC\). That angle is \(\angle ABC\). Therefore \(\angle ABC=37^\circ\).

Equal chords \(AB\) and \(AC\) cut off equal arcs. Angle \(\angle ADB\) stands on chord \(AB\), and angle \(\angle ADC\) stands on chord \(AC\). Therefore these inscribed angles are equal.

We have \(OA=OB\), since these are radii, \(AM=MB\) by condition, and \(OM\) is common. Hence \(\triangle OMA\cong\triangle OMB\). Therefore \(\angle OMA=\angle OMB\). These angles are supplementary, so each is \(90^\circ\). Thus \(OM\perp AB\).

If a point sees segment \(AB\) under a right angle, then it lies on the circle with diameter \(AB\). Since \(\angle ACB=90^\circ\) and \(\angle ADB=90^\circ\), both points \(C\) and \(D\) lie on this circle. Hence \(A,B,C,D\) lie on one circle.

The set of points from which segment \(AB\) is seen under a fixed angle lies on an arc of a circle through \(A\) and \(B\). Since \(C\) and \(D\) are on the same side of \(AB\) and \(\angle ACB=\angle ADB\), they lie on the same such arc. Therefore \(A,B,C,D\) lie on one circle.

The angle between the tangent and \(AB\) equals the inscribed angle standing on \(AB\), namely \(\angle ACB=43^\circ\). The angle between the tangent and \(AC\) equals the inscribed angle standing on \(AC\), namely \(\angle ABC=72^\circ\).

Since \(AB\parallel CD\), we have \(\angle ABD=\angle BDC\). In the cyclic quadrilateral, angles \(\angle BDC\) and \(\angle BAC\) stand on chord \(BC\), so they are equal. Hence \(\angle ABD=\angle BAC\). The first angle stands on chord \(AD\), and the second on chord \(BC\). Equal inscribed angles stand on equal chords, so \(AD=BC\).

Since \(BD\perp AC\) and \(D\in AC\), we get \(\angle BDC=90^\circ\). Since \(CE\perp AB\) and \(E\in AB\), we get \(\angle BEC=90^\circ\). Thus points \(D\) and \(E\) see segment \(BC\) under a right angle. Therefore they lie on the circle with diameter \(BC\), so \(B,C,D,E\) are cyclic.

By the tangent-chord theorem, the angle between the tangent and \(AB\) equals \(\angle ACB\). Since the tangent is parallel to \(BC\), the same angle equals \(\angle ABC\). Hence \(\angle ABC=\angle ACB\), so \(AB=AC\).

Equal chords \(AB\) and \(CD\) cut off equal arcs. Inscribed angles standing on equal arcs are equal. Therefore \(\angle ADB=\angle CAD\).

Since \(OM\perp AB\), point \(M\) is the midpoint of \(AB\). Similarly, \(N\) is the midpoint of \(CD\). From \(AB=CD\), we get \(AM=CN\). In right triangles \(OMA\) and \(ONC\), hypotenuses \(OA\) and \(OC\) are equal as radii, and legs \(AM\) and \(CN\) are equal. Thus the triangles are congruent, so \(OM=ON\).

Angle \(\angle BAC\) stands on chord \(BC\), and angle \(\angle CAD\) stands on chord \(CD\). By condition \(BC=CD\), so the corresponding arcs are equal. Therefore \(\angle BAC=\angle CAD\). Hence \(AC\) is the angle bisector of \(\angle BAD\).

Since \(OA\perp TA\) and \(OB\perp TB\), triangles \(OTA\) and \(OTB\) are right triangles. In them, \(OA=OB\) as radii, and \(OT\) is the common hypotenuse. Hence the triangles are congruent. Therefore \(TA=TB\), and also \(\angle ATO=\angle OTB\), so \(OT\) bisects angle \(ATB\).

Points \(A,D,B\) are collinear, so \(\angle BDE=180^\circ-\angle ADE\). By condition \(\angle ADE=\angle ACB\). Since \(E\) lies on \(AC\), angle \(\angle BCE=\angle BCA=\angle ACB\). Thus \(\angle BDE+\angle BCE=180^\circ\). The sum of opposite angles of quadrilateral \(BCDE\) is \(180^\circ\), therefore \(B,C,D,E\) lie on one circle.

Since \(P\) lies on \(AC\), angle \(\angle BAP=\angle BAC\). Since \(P\) lies on \(BD\), angle \(\angle ABP=\angle ABD\). Angle \(\angle APD\) is an exterior angle of triangle \(ABP\) at vertex \(P\). Therefore it equals the sum of the two remote interior angles: \(\angle APD=\angle BAP+\angle ABP=\angle BAC+\angle ABD\).

Since \(BD\perp AC\), we have \(\angle ADH=90^\circ\). Since \(CE\perp AB\), we have \(\angle AEH=90^\circ\). Thus points \(D\) and \(E\) see segment \(AH\) under a right angle, so \(A,D,H,E\) lie on the circle with diameter \(AH\). Also, \(\angle BDC=90^\circ\) and \(\angle BEC=90^\circ\), so \(D\) and \(E\) see segment \(BC\) under a right angle. Therefore \(B,C,D,E\) lie on the circle with diameter \(BC\).

Radii to the points of tangency are perpendicular to the tangents: \(OA\perp TA\) and \(OB\perp TB\). Therefore in quadrilateral \(AOBT\), the angles at \(A\) and \(B\) are \(90^\circ\). The sum of angles in a quadrilateral is \(360^\circ\), so \(\angle ATB=360^\circ-90^\circ-90^\circ-132^\circ=48^\circ\).

For the circle through \(A,B,C\), the tangent at \(B\) forms with chord \(BA\) an angle equal to the inscribed angle \(\angle BCA\), which stands on chord \(BA\). By condition, line \(l\) forms exactly this angle with \(BA\) and lies outside the triangle. Therefore \(l\) is the tangent to the circumcircle at point \(B\).

Angle \(\angle ABD\) stands on chord \(AD\), and angle \(\angle DBC\) stands on chord \(DC\). In one circle, equal inscribed angles stand on equal chords. Since \(\angle ABD=\angle DBC\), we get \(AD=DC\).

Since \(CT\parallel AB\), angle \(\angle ACT\) equals the angle between \(AC\) and \(AB\), that is \(\angle CAB\). By the tangent-chord theorem, \(\angle TAC=\angle ABC\). In triangle \(ACT\), equality \(AT=CT\) is equivalent to equality of base angles: \(\angle ACT=\angle TAC\). Hence \(AT=CT\) if and only if \(\angle CAB=\angle ABC\), that is, if and only if \(AC=BC\).

Since \(BD\perp AC\), line \(HD\) is perpendicular to \(AC\). Since \(CE\perp AB\), line \(HE\) is perpendicular to \(AB\). The angle between two lines equals the angle between their perpendiculars or supplements it to \(180^\circ\), depending on the chosen angle. Inside triangle \(DHE\), the angle is obtuse, so \(\angle DHE=180^\circ-\angle BAC=180^\circ-\angle A\).

Let \(O\) be the circumcentre. Radii \(OB\) and \(OC\) are perpendicular to tangents \(TB\) and \(TC\), so \(\angle OBT=\angle OCT=90^\circ\). The central angle \(\angle BOC\) standing on arc \(BC\) equals \(2\angle BAC\). In quadrilateral \(BOCT\), the angle sum is \(360^\circ\). Hence \(\angle BTC=360^\circ-90^\circ-90^\circ-\angle BOC=180^\circ-2\angle BAC\).

Since \(BD\perp AC\), we have \(\angle ADH=90^\circ\). Since \(CE\perp AB\), we have \(\angle AEH=90^\circ\). Thus \(A,D,H,E\) lie on the circle with diameter \(AH\). In this circle, angles \(\angle ADE\) and \(\angle AHE\) stand on the same chord \(AE\). Therefore \(\angle ADE=\angle AHE\).

E. Full solution.

Since \(PX\perp AB\) and \(PY\perp CB\), we have \(\angle PXB=\angle PYB=90^\circ\). Therefore points \(P,X,B,Y\) lie on one circle with diameter \(PB\).

From this cyclic quadrilateral, \(\angle XYB=\angle XPB\), because both angles subtend chord \(XB\). Thus the angle between \(XY\) and \(CB\) equals the angle between \(XP\) and tangent \(PB\).

Since \(XP\perp AB\), the angle between \(XP\) and \(PB\) equals \(90^\circ-\angle ABP\). By the tangent-chord theorem, \(\angle ABP=\angle ACB\).

Hence the angle between \(XY\) and \(CB\) is \(90^\circ-\angle ACB\). This exactly means that \(XY\perp AC\).

E. Full solution.

Assume the three initial points were not collinear. Then they lie on a unique circle \(\omega\).

We prove by induction that every newly marked point also lies on \(\omega\). Suppose chosen points \(A,B,C\) already lie on \(\omega\), and \(D\) is obtained by reflecting \(A\) across the perpendicular bisector of \(BC\).

The perpendicular bisector of chord \(BC\) passes through the center of circle \(\omega\). Hence \(\omega\) is symmetric with respect to this line. Therefore the reflection of point \(A\), which lies on \(\omega\), also lies on \(\omega\). Thus \(D\in\omega\).

So all marked points lie on one circle. But a line can meet a circle in at most two points, unless it coincides with the circle, which is impossible. Therefore three distinct marked points could not become collinear. This is a contradiction.

Hence, if three collinear points do appear, the three initial points must have been collinear.

E. Full solution.

Since \(PB\) and \(PC\) are tangents to \(\Omega\), radii \(OB\) and \(OC\) are perpendicular to them. Hence \(\angle OBP=\angle OCP=90^\circ\).

Thus points \(O,B,C,P\) lie on one circle, and \(OP\) is its diameter. Since \(S\) lies on this circle, \(\angle OSP=90^\circ\).

On the other hand, \(S\) lies on the circle with diameter \(AO\). Therefore \(\angle ASO=90^\circ\).

So both lines \(SP\) and \(SA\) pass through point \(S\) and are perpendicular to \(OS\). Through a point there is only one perpendicular to a given line; hence \(A,S,P\) are collinear.

E. Full solution.

On the extension of \(BC\) beyond \(C\), mark point \(N\) so that \(CN=CL\). Then \(BN=BC+CN=BC+CL\), so it is enough to prove \(BN=AB\).

Since \(A,B,K,L\) lie on one circle, angles \(\angle AKB\) and \(\angle ALB\) are supplementary. Using the collinearities \(A,K,C\) and \(B,L,C,N\), we get \(\angle CKB=\angle ALC\).

Triangles \(ACL\) and \(ACN\) are right triangles, have common leg \(AC\), and have equal legs \(CL=CN\). They are congruent, hence \(\angle ANC=\angle ALC\). Therefore \(\angle ANB=\angle CKB\).

Since \(BK\) bisects angle \(B\), \(\angle ABK=\angle KBC\). Using the right triangle and the previous angle equality, we obtain \(\angle BAN=\angle ANB\).

Thus triangle \(ABN\) is isosceles, and \(AB=BN\). Therefore \(AB=BC+CL\).

E. Full solution.

Since \(BH\perp AC\) and \(K\) lies on ray \(HB\), we have \(KH\perp AH\). Also, by construction, \(KM\perp AM\).

Thus \(\angle AHK=\angle AMK=90^\circ\), so points \(A,H,M,K\) lie on one circle. This circle has diameter \(AK\).

Angle \(\angle MAH\) equals \(\angle MAC=30^\circ\), because \(H\in AC\). Hence chord \(HM\), subtending a \(30^\circ\) angle, equals the radius of the circle, so \(HM=\frac{AK}{2}\).

In right triangle \(BHC\), point \(M\) is the midpoint of hypotenuse \(BC\), because \(AM\) is a median of triangle \(ABC\). Therefore \(HM=\frac{BC}{2}\).

We get \(\frac{AK}{2}=HM=\frac{BC}{2}\), hence \(AK=BC\).

By the area formula, \(S=\frac{1}{2}\cdot18\cdot7=63\).

Both triangles have the height from point \(A\) to line \(BC\). Therefore their areas are in the ratio of their bases on this line: \(S_{ABD}:S_{ACD}=BD:DC\).

Since \(M\) is the midpoint of \(BC\), \(BM=CM\). Triangles \(ABM\) and \(ACM\) have the common height from \(A\) to line \(BC\). Therefore their areas are equal.

Triangles \(ABC\) and \(ABD\) have common base \(AB\). The heights from \(C\) and \(D\) to \(AB\) are equal because these points lie on a line parallel to \(AB\). Hence the areas are equal.

Let \(ABCD\) be a parallelogram. Diagonal \(AC\) divides it into triangles \(ABC\) and \(ACD\). They have common base \(AC\), and the heights from \(B\) and \(D\) to \(AC\) are equal because \(AB\parallel CD\). Hence these triangle areas are equal.

Since \(BD:DC=3:4\), we have \(BD:BC=3:7\). Triangles \(ABD\) and \(ABC\) have the common height from \(A\) to \(BC\). Therefore \(S_{ABD}:S_{ABC}=BD:BC=3:7\).

Since \(CD\parallel AB\), points \(C\) and \(D\) lie on one line parallel to base \(AB\). Therefore the heights to base \(AB\) are equal. The base is common, so the areas are equal.

Median \(AM\) divides the triangle into two equal-area triangles. Therefore \(S_{ABM}=\frac{84}{2}=42\).

We have \(BD=\frac{1}{2}BA\) and \(BE=\frac{1}{2}BC\), and the angle at \(B\) is common. The area is reduced by \(2\cdot2=4\) times. Therefore \(S_{BDE}=\frac{64}{4}=16\).

Triangles \(ABC\) and \(ACD\) have bases \(BC\) and \(AD\) on parallel lines, and the heights to these bases equal the distance between the bases of the trapezoid. Therefore \(S_{ABC}:S_{ACD}=BC:AD=9:15=3:5\).

Triangles \(ABD\) and \(ACD\) have the common height from \(A\). Therefore \(BD:DC=S_{ABD}:S_{ACD}=24:40=3:5\).

Since base \(AB\) is common and the areas are equal, the heights from \(C\) and \(D\) to \(AB\) are equal. Points on the same side of \(AB\) at the same distance from \(AB\) lie on a line parallel to \(AB\). Hence \(CD\parallel AB\).

Denote the areas of \(AGF\), \(BGF\), \(BGD\), \(CGD\), \(CGE\), \(AGE\) in order by \(x_1,x_2,x_3,x_4,x_5,x_6\). Since \(F,D,E\) are side midpoints, we get \(x_1=x_2\), \(x_3=x_4\), \(x_5=x_6\). Median \(AD\) halves the whole triangle, so \(x_1+x_2+x_3=x_4+x_5+x_6\). Median \(BE\) gives \(x_1+x_2+x_6=x_3+x_4+x_5\). Substituting the pairwise equalities, we get \(x_1=x_5\) and \(x_1=x_3\). Hence all six areas are equal.

Since \(DE\parallel BC\), \(\triangle ADE\sim\triangle ABC\). We have \(AD:AB=2:5\). Therefore the area ratio equals the square of the side ratio: \(S_{ADE}:S_{ABC}=4:25\).

Triangles \(ABD\), \(ADE\), \(AEC\) have bases \(BD\), \(DE\), \(EC\) on one line \(BC\) and a common height from \(A\). Since the bases are equal, the areas are equal.

Let \(AP:PC=t:(1-t)\). In triangle \(ABC\), \(S_{ABP}:S_{PBC}=AP:PC\), so \(S_{ABP}=tS_{ABC}\). Similarly, in triangle \(ACD\), \(S_{ADP}=tS_{ACD}\). But a diagonal of a parallelogram divides it into two equal-area triangles: \(S_{ABC}=S_{ACD}\). Hence \(S_{ABP}=S_{ADP}\).

Since \(DE\parallel AC\), \(\triangle BDE\sim\triangle BCA\). We have \(BD:BC=2:5\). Therefore areas are in the ratio of squares of corresponding sides: \(S_{BDE}:S_{ABC}=4:25\).

Since \(P\) lies on \(AM\), line \(AP\) is the median and passes through the midpoint \(M\) of side \(BC\). Triangles \(PAB\) and \(PAC\) have common base \(AP\). Points \(B\) and \(C\) are at equal distances from line \(AP\), so the areas are equal.

Triangles \(PAB\) and \(PAC\) have common base \(AP\). Equality of their areas implies that the distances from \(B\) and \(C\) to line \(AP\) are equal. Since \(B\) and \(C\) lie on opposite sides of \(AP\), line \(AP\) passes through the midpoint of \(BC\). Thus \(AP\) is a median.

Triangles \(BCE\) and \(BCA\) have common base \(BC\). Since \(AE:ED=1:2\), point \(E\) is \(\frac{1}{3}\) of the way from \(A\) to \(D\), where \(D\) lies on \(BC\). Therefore the height from \(E\) to \(BC\) is \(\frac{2}{3}\) of the height from \(A\). Hence \(S_{BCE}=\frac{2}{3}\cdot90=60\).

From \(S_{PAB}=S_{PCA}\), line \(AP\) passes through the midpoint of \(BC\), so \(AP\) is a median. From \(S_{PAB}=S_{PBC}\), line \(BP\) passes through the midpoint of \(AC\), so \(BP\) is a median. Therefore \(P\) is the intersection of two medians, hence the centroid of the triangle.

Triangles \(BCE\) and \(BCA\) have common base \(BC\). The height of point \(E\) to \(BC\) is \(ED:AD\) of the height of point \(A\), because \(D\) lies on \(BC\). From \(AE:ED=3:2\), we get \(ED:AD=2:5\). Therefore \(S_{BCE}=\frac{2}{5}\cdot120=48\).

Let the angle between the diagonals be \(\varphi\). Then \(S_{AOB}=\frac{1}{2}AO\cdot BO\sin\varphi\), \(S_{COD}=\frac{1}{2}CO\cdot DO\sin\varphi\). Similarly, \(S_{BOC}=\frac{1}{2}BO\cdot CO\sin\varphi\), \(S_{DOA}=\frac{1}{2}DO\cdot AO\sin\varphi\). The products on the left and right are identical, so the equality is proved.

Triangles \(ABD\) and \(ACD\) have common base \(AD\). The heights from \(B\) and \(C\) to \(AD\) are equal because \(BC\parallel AD\). Therefore \(S_{ABD}=S_{ACD}\). Subtract the common triangle \(AOD\) from both areas: we get \(S_{AOB}=S_{COD}\).

Triangles \(BDE\) and \(BCA\) are similar because \(DE\parallel AC\). The similarity ratio is \(BD:BC=2:5\). Therefore \(S_{BDE}:S_{ABC}=4:25\). The remaining part \(ADEC\) has area \(21\) parts out of \(25\). Hence \(S_{BDE}:S_{ADEC}=4:21\).

Let the distances from \(P\) to lines \(AB\) and \(CD\) be \(h_1\) and \(h_2\). Then \(h_1+h_2\) equals the height of the parallelogram to base \(AB\). Therefore \(S_{PAB}+S_{PCD}=\frac{1}{2}AB\cdot h_1+\frac{1}{2}CD\cdot h_2\). Since \(AB=CD\), this is \(\frac{1}{2}AB(h_1+h_2)=\frac{1}{2}S_{ABCD}\).

Since \(P,A,D\) are collinear, triangles \(ABP\) and \(PBD\) have heights to this line from \(B\), and triangles \(ACP\) and \(PCD\) have heights from \(C\). Comparing fractions on the same line gives \(\frac{BD}{DC}=\frac{S_{ABP}}{S_{ACP}}\). Similarly, \(\frac{CE}{EA}=\frac{S_{BCP}}{S_{ABP}}\) and \(\frac{AF}{FB}=\frac{S_{ACP}}{S_{BCP}}\). Multiplying, we get \(1\).

By the area form of Ceva's theorem, \(\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}=1\). Substitute: \(\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{AF}{FB}=1\). We get \(\frac{1}{2}\cdot\frac{AF}{FB}=1\), hence \(AF:FB=2:1\).

Since \(G\) lies on the median from \(A\), we have \(S_{GAB}=S_{GAC}\). Since \(G\) lies on the median from \(B\), we have \(S_{GAB}=S_{GBC}\). Hence \(S_{GAC}=S_{GBC}\). Triangles \(GAC\) and \(GBC\) have common base \(GC\), so points \(A\) and \(B\) are at equal distances from line \(GC\). Therefore line \(GC\) passes through the midpoint of \(AB\), that is, \(AF=FB\).

We have \(\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}=\frac{2}{3}\cdot\frac{3}{5}\cdot\frac{5}{2}=1\). By the converse form of the area version of Ceva's theorem, product \(1\) implies that lines \(AD\), \(BE\), \(CF\) are concurrent, meaning they meet at one point.

E. Full solution.

Assume such a square has been tiled, and let \(1\times n\) be the longest chosen rectangle. Since the square has area greater than \(1\), we have \(n>1\).

The rectangle \(1\times n\) must fit inside the square, even if rotated. Therefore the side length of the square is at least \(n\), and its area is at least \(n^2\).

On the other hand, the chosen rectangles can only be among \(1\times1,1\times2,\ldots,1\times n\). Their total area is at most \(1+2+\cdots+n=\frac{n(n+1)}{2}\).

For \(n>1\), \(\frac{n(n+1)}{2}

Therefore a square of area greater than \(1\) cannot be tiled in this way.

The diagonals of quadrilateral \(ABDC\) are \(AD\) and \(BC\). By construction \(AM=MD\), and by condition \(BM=MC\). Hence the diagonals are bisected by point \(M\). Therefore \(ABDC\) is a parallelogram.

Since \(DE\parallel BC\), triangles \(ADE\) and \(ABC\) are similar. Since \(D\) is the midpoint of \(AB\), \(AD:AB=1:2\). Hence \(AE:AC=1:2\), so \(AE=EC\). Therefore \(E\) is the midpoint of \(AC\).

The set of points from which segment \(AB\) is seen under a right angle is the circle with diameter \(AB\). Since \(\angle ACB=90^\circ\), point \(C\) lies on this circle.

By construction, \(BD\parallel AC\) and \(CD\parallel AB\). Therefore in quadrilateral \(ABDC\), both pairs of opposite sides are parallel. Hence \(ABDC\) is a parallelogram.

Since \(M\) is the midpoint of \(BC\), \(BM=MC\). Since \(D\) is the reflection of \(A\) about \(M\), \(AM=MD\). Thus point \(M\) bisects diagonals \(BC\) and \(AD\) of quadrilateral \(ABDC\).

Since \(D\) lies on ray \(BA\), \(\angle DBC=\angle ABC=52^\circ\). In triangle \(BCD\), sides \(BD\) and \(BC\) are equal, so \(\angle BCD=\angle CDB\). Hence \(\angle BCD=\frac{180^\circ-52^\circ}{2}=64^\circ\).

Since \(DE\parallel AB\), triangles \(CDE\) and \(CBA\) are similar. We have \(CD:CB=3:5\), hence \(CE:CA=3:5\). Therefore \(CE:EA=3:2\).

Draw the circle with diameter \(AB\). Since \(\angle ACB=90^\circ\), point \(C\) lies on this circle. Since \(\angle ADB=90^\circ\), point \(D\) also lies on this circle. Therefore \(A,B,C,D\) lie on one circle.

By construction, \(ABDC\) is a parallelogram. In a parallelogram, the diagonals bisect each other. Here the diagonals are \(AD\) and \(BC\). Therefore the intersection of \(AD\) and \(BC\) is the midpoint of \(BC\).

Draw the circle through \(A,B,C\). On one side of \(AB\), all points from which segment \(AB\) is seen under angle \(\angle ACB\) lie on one arc of this circle. Since \(\angle ADB=\angle ACB\), point \(D\) lies on this circle. Hence \(A,B,C,D\) are cyclic.

Construct point \(P\) on the extension of \(MN\) so that \(NP=MN\). In triangles \(AMN\) and \(CNP\), we have \(AN=NC\), \(MN=NP\), and the angles at \(N\) are vertical. Hence the triangles are congruent, so \(AM=CP\) and \(AM\parallel CP\). Since \(AM=MB\), we get \(CP=MB\) and \(CP\parallel MB\), so \(BMCP\) is a parallelogram. Therefore \(MP\parallel BC\), hence \(MN\parallel BC\).

We have \(AM=MD\) by construction, \(BM=MC\) because \(M\) is the midpoint of \(BC\), and \(\angle AMB=\angle DMC\) as vertical angles. Therefore \(\triangle ABM\cong\triangle DCM\) by two sides and the included angle.

As in the basic construction, \(ABDC\) is a parallelogram. Therefore \(BD=AC\) and \(CD=AB\). By condition \(AB=AC\), hence \(BD=CD\).

Since \(DE\parallel AC\), triangles \(BDE\) and \(BAC\) are similar. From \(BD:BA=1:2\), it follows that \(BE:BC=1:2\). Hence \(BE=EC\), so \(E\) is the midpoint of \(BC\).

Since \(E\) lies on \(AC\), side \(AE\) is parallel to the line through \(D\) parallel to \(AC\), namely \(DF\). Since \(F\) lies on \(AB\), side \(AF\) is parallel to the line through \(D\) parallel to \(AB\), namely \(DE\). Thus \(AE\parallel DF\) and \(AF\parallel DE\), so \(AEDF\) is a parallelogram.

Since \(BD\perp AC\), and \(D\in AC\), we get \(\angle BDC=90^\circ\). Since \(CE\perp AB\), and \(E\in AB\), we get \(\angle BEC=90^\circ\). Hence points \(D\) and \(E\) lie on the circle with diameter \(BC\). Therefore \(B,C,D,E\) lie on one circle.

We have \(AM=MD\), \(BM=MC\), and \(\angle AMB=\angle DMC\) as vertical angles. Thus \(\triangle ABM\cong\triangle DCM\). Therefore \(\angle ABM=\angle DCM\). These are alternate interior angles for lines \(AB\) and \(CD\) with transversal \(BC\), so \(AB\parallel CD\).

Opposite sides of a parallelogram are equal. Therefore in \(ABDC\), \(AB=CD\) and \(AC=BD\). This is why a completed parallelogram often lets us move a segment to a convenient place.

Angles \(\angle ABC\) and \(\angle ADC\) are inscribed and stand on the same chord \(AC\). Therefore they are equal. In similar problems, a circle is useful because it lets us replace one angle by another angle located in a more convenient part of the diagram.