Difference between revisions of "1974 USAMO Problems/Problem 5"
(Problem and solution) |
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</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>: | ||
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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 01:15, 3 January 2009
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>