Triangles II: Similarity

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 \(C\).