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Difference between revisions of "1974 USAMO Problems/Problem 5"

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== See Also ==
 
== See Also ==
  
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{{USAMO box|year=1974|num-b=4|aftertext=|after=Last Problem}}
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{{USAMO box|year=1974|num-b=4|aftertext=|after=Last Question}}
  
 
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*<url>viewtopic.php?t=33494 Discussion on AoPS/MathLinks</url>
  
 
[[Category:Olympiad Geometry Problems]]
 
[[Category:Olympiad Geometry Problems]]

Revision as of 14:19, 17 September 2012

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 $x=u+v+w$. [asy] size(400); defaultpen(1); pair C=(0,1), A=(1,6), B=(4,1), D=(1.5,2); draw(D--A--B--C--D--B); draw(A--C); label("$A$",A,N); label("$B$",B,ESE); label("$C$",C,SW); label("$D$",D,NE); label("$a$",(B+C)/2,S); label("$b$",(C+A)/2,WNW); label("$c$",(A+B)/2,NE); label("$u$",(A+D)/2,E); label("$v$",(B+D)/2,N); label("$w$",(C+D)/2,N); pair P=(7,0), Q=(14,0), R=P+7expi(pi/3), M=(10,1.2); draw(M--P--Q--R--M--Q); draw(P--R); label("$P$",P,W); label("$Q$",Q,E); label("$R$",R,W); label("$M$",M,NE); label("$x$",(P+Q)/2,S); label("$x$",(Q+R)/2,E); label("$x$",(R+P)/2,W); label("$a$",(P+M)/2,SE); label("$b$",(Q+M)/2,N); label("$c$",(R+M)/2,E); label("Figure 1",P-(0,1),S); [/asy]

Solutions

Solution 1

We rotate figure $PRQM$ by a clockwise angle of $\pi/3$ about $Q$ to obtain figure $RR'QM'$: [asy] size(300); defaultpen(1); pair P=(7,0), Q=(14,0), R=P+7expi(pi/3), M=(10,1.2); pair RR=R+Q-P, MM= rotate(-60,Q)*M; draw(P--R--RR--Q--P--M--MM--RR); draw(Q--R--M--Q--MM--R); label("$P$",P,W); label("$Q$",Q,E); label("$R$",R,W); label("$M$",M,NW); label("$R'$",RR,NE); label("$M'$",MM,ESE); [/asy]

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 \bigr] \sqrt{3}}{4} \\ &= 3 [ABC] + \frac{(a^2+b^2+c^2)\sqrt{3}}{4} = 2[PQR] . \end{align*} 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 $A+60$. [asy] size(200); defaultpen(fontsize(8)); pair A=6*expi(5/9*pi), B=3*expi(5/9*pi+2*pi/3), C=4*expi(5/9*pi+4*pi/3), D=(0,0); path triabc = D--A--B--C--D--B..A--C; draw(triabc); label("$A$",A,( 0, 1)); label("$B$",B,(-1,-1));label("$b$",(C+A)/2,( 1, 1)); label("$C$",C,( 1,-1));label("$w$",(C+D)/2,( 0, 1)); label("$D$",D,( 1, 1)); transform rot60a = rotate(-60,A); pair A1 = rot60a*A, B1 = rot60a*B, C1 = rot60a*C, D1 = rot60a*D; path triabc1 = D1--A1--B1--C1--D1--B1..A1--C1; draw(triabc1, linetype("8 8")); label("$B'$",B1,(-1, 0));label("$v$",(B1+D1)/2,( 0, 1)); label("$C'$",C1,( 0,-1));label("$c$",(A1+B1)/2,( 0, 1)); label("$D'$",D1,(-1,-1));label("$u$",(D1+D)/2,( 0, -1)); draw(A--B1--C--A, red+1.5); dot(A^^B1^^D1^^D^^C); [/asy]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 $C+60$.[asy] size(400); defaultpen(fontsize(8)); picture transC; pair A=6*expi(5/9*pi), B=3*expi(5/9*pi+2*pi/3), C=4*expi(5/9*pi+4*pi/3), D=(0,0); path triabc = D--A--B--C--D--B..A--C; draw(triabc); label("$A$",A,( 0, 1));label("$u$",(A+D)/2,( 1, 0)); label("$B$",B,(-1,-1)); label("$C$",C,( 1,-1));label("$c$",(A+B)/2,(-1, 1)); label("$D$",D,( 1, 1)); transform rot60b = rotate(-60,B); pair A1 = rot60b*A, B1 = rot60b*B, C1 = rot60b*C, D1 = rot60b*D; path triabc1 = D1--A1--B1--C1--D1--B1..A1--C1; draw(triabc1, linetype("8 8")); label("$A'$",A1,( 1, 0));label("$a$",(B1+C1)/2,(-1, 0)); label("$C'$",C1,( 0,-1));label("$w$",(C1+D1)/2,( 1, 0)); label("$D'$",D1,(-1,-1));label("$v$",(D1+D)/2,( 1, 1)); draw(A--B--C1--A, red+1.5); dot(A^^B^^D1^^D^^C1); draw(transC, triabc); label(transC, "$A$",A,( 0, 1));label(transC, "$a$",(B+C)/2,( 0,-1)); label(transC, "$B$",B,(-1,-1));label(transC, "$v$",(B+D)/2,( 0, 1)); label(transC, "$C$",C,( 1,-1)); label(transC, "$D$",D,( 0,-1)); transform rot60c = rotate(-60,C); pair A2 = rot60c*A, B2 = rot60c*B, C2 = rot60c*C, D2 = rot60c*D; path triabc2 = D2--A2--B2--C2--D2--B2..A2--C2; draw(transC, triabc2, linetype("8 8")); label(transC, "$A'$",A2,( 1, 0));label(transC, "$u$",(A2+D2)/2,( 1,-1)); label(transC, "$b$",(C2+A2)/2,( 1,-1)); label(transC, "$w$",(D+D2)/2,( 1,-1)); label(transC, "$D'$",D2,( 0, 1)); draw(transC, A2--B--C--A2, red+1.5); dot(transC, C^^B^^D2^^D^^A2); add(shift(15*right)*transC); [/asy] These three triangles fit together because $(A+60)+(B+60)+(C+60) = 360$. The result is an equilateral triangle of side length $u+v+w$.

Solution 3

[asy] size(300); defaultpen(1); pair P=(7,0), Q=(14,0), R=P+7expi(pi/3), M=(10,1.2); pair RR=R+Q-P, MM= rotate(-60,Q)*M; draw(P--R--RR--Q--P--M--MM--RR); draw(Q--R--M--Q--MM--R); label("$P$",P,W); label("$Q$",Q,E); label("$R$",R,W); label("$M$",M,NW); label("$R'$",RR,NE); label("$M'$",MM,ESE); [/asy]

As in the first solution, we rotate and establish that $\triangle MRM' \cong \triangle ABC$.

Let $X$ and $Y$ be points on $\overline{RQ}$ and $\overline{PQ}$, respectively, such that $M$ lies on $\overline{XY}$ and $\overline{XY}\parallel \overline{PR}$. We note that $m\angle RXM = 120^\circ$. The rotation then takes $Y$ to $X$, so $m\angle RXM'=m\angle PYM = 120^\circ$. It follows that $RX=BD=v$, $MX=AD=u$, $M'X=CD=w$.

Since $m\angle MXM' = 120^\circ$ and $m\angle MQM' = 60^\circ$, $MXM'Q$ is cyclic. By Ptolemy's theorem, \begin{align*} (MX)(M'Q) + (XM')(QM) &= (MM')(XQ) \\ XQ &= MX + M'X \\ &= u+w. \end{align*} Finally, $RQ = RX+XQ = u+v+w$, as desired.



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

See Also

1974 USAMO (Problems • Resources)
Preceded by
Problem 4 		Last Question
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions
  • <url>viewtopic.php?t=33494 Discussion on AoPS/MathLinks</url>
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