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1974 USAMO Problems/Problem 5

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Contents

1 Problem
2 Solutions
- 2.1 Solution 1
- 2.2 Solution 2
3 Resources

Problem

Consider the two triangles $\triangle ABC$ and $\triangle PQR$ shown in Figure 1. In $\triangle ABC$ , $\angle ADB = \angle BDC = \angle CDA = 120^\circ$ . Prove that $x=u+v+w$ .

Solutions

Solution 1

We rotate figure $PRQM$ by a clockwise angle of $\pi/3$ about $Q$ to obtain figure $RR'QM'$ :

Evidently, $MM'Q$ is an equilateral triangle, so triangles $MRM'$ and $ABC$ are congruent. Also, triangles $PMQ$ and $RM'Q$ are congruent, since they are images of each other under rotations. Then $[ABC] + \frac{b^2 \sqrt{3}}{4} = [MRM'] + [MM'Q] = [QMR] + [RM'Q] = [QMR] + [PMQ] .$ Then by symmetry, $3 [ABC] + \frac{(a^2+b^2+c^2)\sqrt{3}}{4} = 2\bigl( [PMQ] + [QMR] + [RMP] \bigr) = 2 [PRQ] .$

But $ABC$ is composed of three smaller triangles. The one with sides $w,v,a$ has area $\tfrac{1}{2} wv \sin 120^\circ = \frac{wv \sqrt{3}}{4}$ . Therefore, the area of $ABC$ is $\frac{(wv+vu+uw)\sqrt{3}}{4} .$ Also, by the Law of Cosines on that small triangle of $ABC$ , $a^2 = w^2 + wv+ v^2$ , so by symmetry, $\frac{(a^2 + b^2 + c^2)\sqrt{3}}{4} = \frac{\bigl[2(u^2+w^2+v^2) + wv + vu + uw \bigr] \sqrt{3}}{4}.$ Therefore $\begin{align*} \frac{(u+v+w)^2 \sqrt{3}}{2} &= 3 \frac{(wv+vu+uw)\sqrt{3}}{4} + \frac{\bigl[2(u^2+w^2+v^2) + wv + vu + uw...$ But the area of triangle $PQR$ is $x^2 \sqrt{3}/4$ . It follows that $u+v+w=x$ , as desired. $\blacksquare$

Solution 2

Rotate $\triangle ABC$ $60$ degrees clockwise about $A$ to get $\triangle AB'C'$ . Observe that $\triangle ADD'$ is equilateral, which means $D'D=AD=u$ . Also, $B',D',D,C$ are collinear because $\angle B'D'A + \angle AD'D = 120+60=180$ and $\angle CDA + \angle DD'A = 120+60=180$ . The resulting $\triangle B'AC$ has side lengths $b,c,u+v+w$ and the angle opposite side $u+v+w$ has magnitude $A+60$ . If we perform the rotation about points $B$ and $C$ , we get two triangles. One has side lengths $a,c,u+v+w$ and the angle opposite side $u+v+w$ has magnitude $B+60$ , and the other has side lengths $b,c,u+v+w$ and the angle opposite side $u+v+w$ has magnitude $C+60$ . These three triangles fit together because $(A+60)+(B+60)+(C+60) = 360$ . The result is an equilateral triangle of side length $u+v+w$ .

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

Resources

1974 USAMO (Problems)
Preceded by Problem 4	1 • 2 • 3 • 4 • 5	Last Problem
All USAMO Problems and Solutions

Discussion on AoPS/MathLinks

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Category: Olympiad Geometry Problems

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