Difference between revisions of "2000 AIME I Problems/Problem 14"
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draw(A--B--C--A, black+1);draw(B--P--Q); MP("x",B,20*dir(75),p); MP("x",P,17*dir(245),p); MP("2x",Q,15*dir(70),p); MP("2x",A,15*dir(-90),p); MP("2y",P,2*left,p); MP("3x",P,10*dir(-95),p); MP("x+y",C,5*dir(135),p); MP("y",B,5*dir(40),p); | draw(A--B--C--A, black+1);draw(B--P--Q); MP("x",B,20*dir(75),p); MP("x",P,17*dir(245),p); MP("2x",Q,15*dir(70),p); MP("2x",A,15*dir(-90),p); MP("2y",P,2*left,p); MP("3x",P,10*dir(-95),p); MP("x+y",C,5*dir(135),p); MP("y",B,5*dir(40),p); | ||
</asy></center> | </asy></center> | ||
− | Let <math>\angle QPB=x^\circ</math>. Because <math>\angle AQP</math> is exterior to isosceles triangle <math>PQB</math> its measure is <math>2x</math> and <math>\angle PAQ</math> has the same measure. Because <math>\angle BPC</math> is exterior to <math>\triangle BPA</math> its measure is <math>3x</math>. Let <math>\angle PBC = y^\circ</math>. It follows that <math>\angle ACB = x+y</math> and that <math>4x+2y=180^\circ</math>. Two of the angles of triangle <math>APQ</math> have measure <math>2x</math>, and thus the measure of <math>\angle APQ</math> is <math>2y</math>. It follows that <math>AQ=2\cdot AP\cdot \sin y</math>. Because <math>AB=AC</math> and <math>AP=QB</math>, it also follows that <math>AQ=PC</math>. Now apply the Law of Sines to triangle <math>PBC</math> to find <cmath>\frac{\sin 3x}{BC}=\frac{\sin y}{PC}=\frac{\sin y}{2\cdot AP\cdot \sin y}= \frac{1}{2\cdot BC}</cmath>because <math>AP=BC</math>. Hence <math>\sin 3x = \ | + | Let <math>\angle QPB=x^\circ</math>. Because <math>\angle AQP</math> is exterior to isosceles triangle <math>PQB</math> its measure is <math>2x</math> and <math>\angle PAQ</math> has the same measure. Because <math>\angle BPC</math> is exterior to <math>\triangle BPA</math> its measure is <math>3x</math>. Let <math>\angle PBC = y^\circ</math>. It follows that <math>\angle ACB = x+y</math> and that <math>4x+2y=180^\circ</math>. Two of the angles of triangle <math>APQ</math> have measure <math>2x</math>, and thus the measure of <math>\angle APQ</math> is <math>2y</math>. It follows that <math>AQ=2\cdot AP\cdot \sin y</math>. Because <math>AB=AC</math> and <math>AP=QB</math>, it also follows that <math>AQ=PC</math>. Now apply the Law of Sines to triangle <math>PBC</math> to find <cmath>\frac{\sin 3x}{BC}=\frac{\sin y}{PC}=\frac{\sin y}{2\cdot AP\cdot \sin y}= \frac{1}{2\cdot BC}</cmath>because <math>AP=BC</math>. Hence <math>\sin 3x = \tfrac 12</math>. Since <math>4x<180</math>, this implies that <math>3x=30</math>, i.e. <math>x=10</math>. Thus <math>y=70</math> and <cmath>r=\frac{10+70}{2\cdot 70}=\frac 47,</cmath>which implies that <math>1000r = 571 + \tfrac 37</math>. So the answer is <math>\boxed{571}</math>. |
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== Solution 1 == | == Solution 1 == |
Revision as of 17:50, 25 January 2022
Problem
In triangle it is given that angles and are congruent. Points and lie on and respectively, so that Angle is times as large as angle where is a positive real number. Find .
Contents
[hide]Official Solution (MAA)
Let . Because is exterior to isosceles triangle its measure is and has the same measure. Because is exterior to its measure is . Let . It follows that and that . Two of the angles of triangle have measure , and thus the measure of is . It follows that . Because and , it also follows that . Now apply the Law of Sines to triangle to find because . Hence . Since , this implies that , i.e. . Thus and which implies that . So the answer is .
Solution 1
Let point be in such that . Then is a rhombus, so and is an isosceles trapezoid. Since bisects , it follows by symmetry in trapezoid that bisects . Thus lies on the perpendicular bisector of , and . Hence is an equilateral triangle.
Now , and the sum of the angles in is . Then and , so the answer is .
Solution 2
Again, construct as above.
Let and , which means . is isosceles with , so . Let be the intersection of and . Since , is cyclic, which means . Since is an isosceles trapezoid, , but since bisects , .
Therefore we have that . We solve the simultaneous equations and to get and . , , so . .
Solution 3 (Trig identities)
Let and . is isosceles, so and . is isosceles too, so . Using the expression for , we get by the triple angle formula! It follows now that , , giving .
The answer is .
See also
2000 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.