Difference between revisions of "2000 AIME I Problems/Problem 14"
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Let the measure of <math>\angle BAC</math> be <math>\alpha</math> and <math>\overline{AP}=\overline{PQ}=\overline{QB}=\overline{BC}=x</math>. Because <math>\triangle APQ</math> is isosceles, <math>AQ=2x\cos(\alpha)</math>. So, <math>\overline{AB}=x\left(2\cos(\alpha)+1\right)</math>. <math>\triangle{ABC}</math> is isosceles too, so <math>x=\overline{BC}=2\overline{AC}\sin\left(\frac{\alpha}{2}\right)=2x\left(2\cos(\alpha)+1\right)\sin\left(\frac{\alpha}{2}\right)</math>. | Let the measure of <math>\angle BAC</math> be <math>\alpha</math> and <math>\overline{AP}=\overline{PQ}=\overline{QB}=\overline{BC}=x</math>. Because <math>\triangle APQ</math> is isosceles, <math>AQ=2x\cos(\alpha)</math>. So, <math>\overline{AB}=x\left(2\cos(\alpha)+1\right)</math>. <math>\triangle{ABC}</math> is isosceles too, so <math>x=\overline{BC}=2\overline{AC}\sin\left(\frac{\alpha}{2}\right)=2x\left(2\cos(\alpha)+1\right)\sin\left(\frac{\alpha}{2}\right)</math>. | ||
− | Simplifying, <math>1=4\cos(\alpha)\sin\left(\frac{\alpha}{2}\right)+ | + | Simplifying, <math>1=4\cos(\alpha)\sin\left(\frac{\alpha}{2}\right)+2\sin\left(\frac{\alpha}{2}\right)</math>. By double angle formula, we know that <math>\cos(\alpha)=1-2\sin^2\left(\frac{\alpha}{2}\right)</math>. |
Applying, <math>4\sin\left(\frac{\alpha}{2}\right)-8\sin^3\left(\frac{\alpha}{2}\right)+2\sin\left(\frac{\alpha}{2}\right)=1</math> and <math>2\left(3\sin\left(\frac{\alpha}{2}\right)-4\sin^3\left(\frac{\alpha}{2}\right)\right)=1</math>. | Applying, <math>4\sin\left(\frac{\alpha}{2}\right)-8\sin^3\left(\frac{\alpha}{2}\right)+2\sin\left(\frac{\alpha}{2}\right)=1</math> and <math>2\left(3\sin\left(\frac{\alpha}{2}\right)-4\sin^3\left(\frac{\alpha}{2}\right)\right)=1</math>. | ||
The expression in the parentheses though is triple angle formula! Hence, <math>\sin\left(\frac{3\alpha}{2}\right)=\frac{1}{2}</math>, <math>\alpha=20^o</math>. It follows now that <math>\angle APQ=140^o</math>, <math>\angle ACB=80^o</math>. Giving <math>r=\frac{4}{7}</math>. | The expression in the parentheses though is triple angle formula! Hence, <math>\sin\left(\frac{3\alpha}{2}\right)=\frac{1}{2}</math>, <math>\alpha=20^o</math>. It follows now that <math>\angle APQ=140^o</math>, <math>\angle ACB=80^o</math>. Giving <math>r=\frac{4}{7}</math>. | ||
<math>\left\lfloor 1000\left(\frac {4}{7}\right)\right\rfloor = \boxed{571}</math>. | <math>\left\lfloor 1000\left(\frac {4}{7}\right)\right\rfloor = \boxed{571}</math>. | ||
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== See also == | == See also == |
Revision as of 21:54, 12 October 2020
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 .
Solution
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
Let the measure of be and . Because is isosceles, . So, . is isosceles too, so . Simplifying, . By double angle formula, we know that . Applying, and . The expression in the parentheses though is triple angle formula! Hence, , . It follows now that , . Giving . .
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.