Difference between revisions of "1974 USAMO Problems/Problem 5"
(Problem and solution) |
|||
Line 38: | Line 38: | ||
</asy> | </asy> | ||
− | == Solution == | + | == Solutions == |
+ | |||
+ | === Solution 1 === | ||
We rotate figure <math>PRQM</math> by a clockwise angle of <math>\pi/3</math> about <math>Q</math> to obtain figure <math>RR'QM'</math>: | We rotate figure <math>PRQM</math> by a clockwise angle of <math>\pi/3</math> about <math>Q</math> to obtain figure <math>RR'QM'</math>: | ||
Line 74: | Line 76: | ||
But the area of triangle <math>PQR</math> is <math>a^2 \sqrt{3}/4</math>. It follows that <math>u+v+w=a</math>, as desired. <math>\blacksquare</math> | But the area of triangle <math>PQR</math> is <math>a^2 \sqrt{3}/4</math>. It follows that <math>u+v+w=a</math>, as desired. <math>\blacksquare</math> | ||
+ | === Solution 2 === | ||
+ | Rotate <math>\triangle ABC</math> <math>60</math> degrees clockwise about <math>A</math> to get <math>\triangle AB'C'</math>. Observe that <math>\triangle ADD'</math> is equilateral, which means <math>D'D=AD=u</math>. Also, <math>B',D',D,C</math> are collinear because <math>\angle B'D'A + \angle AD'D = 120+60=180</math> and <math>\angle CDA + \angle DD'A = 120+60=180</math>. The resulting <math>\triangle B'AC</math> has side lengths <math>b,c,u+v+w</math> and the angle opposite side <math>u+v+w</math> has magnitude <math>A+60</math>. | ||
+ | <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 <math>B</math> and <math>C</math>, we get two triangles. One has side lengths <math>a,c,u+v+w</math> and the angle opposite side <math>u+v+w</math> has magnitude <math>B+60</math>, and the other has side lengths <math>b,c,u+v+w</math> and the angle opposite side <math>u+v+w</math> has magnitude <math>C+60</math>.<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 <math>(A+60)+(B+60)+(C+60) = 360</math>. The result is an equilateral triangle of side length <math>u+v+w</math>. | ||
{{alternate solutions}} | {{alternate solutions}} |
Revision as of 00:15, 3 January 2009
Contents
[hide]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 .
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 • Resources) | ||
Preceded by Problem 4 |
Last Problem | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
- <url>viewtopic.php?t=33494 Discussion on AoPS/MathLinks</url>