Triangles I: Congruence

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.