Difference between revisions of "1974 USAMO Problems/Problem 5"
(→Problem) |
|||
(5 intermediate revisions by 3 users not shown) | |||
Line 75: | Line 75: | ||
&= 3 [ABC] + \frac{(a^2+b^2+c^2)\sqrt{3}}{4} = 2[PQR] . | &= 3 [ABC] + \frac{(a^2+b^2+c^2)\sqrt{3}}{4} = 2[PQR] . | ||
\end{align*} </cmath> | \end{align*} </cmath> | ||
− | But the area of triangle <math>PQR</math> is <math> | + | But the area of triangle <math>PQR</math> is <math>x^2 \sqrt{3}/4</math>. It follows that <math>u+v+w=x</math>, as desired. <math>\blacksquare</math> |
=== Solution 2 === | === Solution 2 === | ||
Line 145: | Line 145: | ||
</asy> | </asy> | ||
These three triangles fit together because <math>(A+60)+(B+60)+(C+60) = 360</math>. The result is an equilateral triangle of side length <math>u+v+w</math>. | These three triangles fit together because <math>(A+60)+(B+60)+(C+60) = 360</math>. The result is an equilateral triangle of side length <math>u+v+w</math>. | ||
+ | |||
+ | ===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 <math>\triangle MRM' \cong \triangle ABC</math>. | ||
+ | |||
+ | Let <math>X</math> and <math>Y</math> be points on <math>\overline{RQ}</math> and <math>\overline{PQ}</math>, respectively, such that <math>M</math> lies on <math>\overline{XY}</math> and <math>\overline{XY}\parallel \overline{PR}</math>. We note that <math>m\angle RXM = 120^\circ </math>. The rotation then takes <math>Y</math> to <math>X</math>, so <math>m\angle RXM'=m\angle PYM = 120^\circ</math>. | ||
+ | It follows that <math>RX=BD=v</math>, <math>MX=AD=u</math>, <math>M'X=CD=w</math>. | ||
+ | |||
+ | Since <math>m\angle MXM' = 120^\circ </math> and <math>m\angle MQM' = 60^\circ </math>, <math>MXM'Q</math> is cyclic. | ||
+ | By Ptolemy's theorem, | ||
+ | <cmath>\begin{align*} | ||
+ | (MX)(M'Q) + (XM')(QM) &= (MM')(XQ) \ | ||
+ | XQ &= MX + M'X \ | ||
+ | &= u+w. | ||
+ | \end{align*}</cmath> | ||
+ | Finally, <math>RQ = RX+XQ = u+v+w</math>, as desired. | ||
+ | |||
+ | |||
+ | |||
{{alternate solutions}} | {{alternate solutions}} | ||
− | == | + | == See Also == |
− | {{USAMO box|year=1974|num-b=4|aftertext=|after=Last | + | {{USAMO box|year=1974|num-b=4|aftertext=|after=Last Question}} |
*<url>viewtopic.php?t=33494 Discussion on AoPS/MathLinks</url> | *<url>viewtopic.php?t=33494 Discussion on AoPS/MathLinks</url> | ||
− | + | {{MAA Notice}} | |
[[Category:Olympiad Geometry Problems]] | [[Category:Olympiad Geometry Problems]] |
Latest revision as of 17:57, 3 July 2013
Problem
Consider the two triangles and shown in Figure 1. In , . Prove that .
Solutions
Solution 1
We rotate figure by a clockwise angle of about to obtain figure :
Evidently, is an equilateral triangle, so triangles and are congruent. Also, triangles and are congruent, since they are images of each other under rotations. Then Then by symmetry,
But is composed of three smaller triangles. The one with sides has area . Therefore, the area of is Also, by the Law of Cosines on that small triangle of , , so by symmetry, Therefore But the area of triangle is . It follows that , as desired.
Solution 2
Rotate degrees clockwise about to get . Observe that is equilateral, which means . Also, are collinear because and . The resulting has side lengths and the angle opposite side has magnitude . If we perform the rotation about points and , we get two triangles. One has side lengths and the angle opposite side has magnitude , and the other has side lengths and the angle opposite side has magnitude . These three triangles fit together because . The result is an equilateral triangle of side length .
Solution 3
As in the first solution, we rotate and establish that .
Let and be points on and , respectively, such that lies on and . We note that . The rotation then takes to , so . It follows that , , .
Since and , is cyclic. By Ptolemy's theorem, Finally, , 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>
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.