Difference between revisions of "2008 AIME I Problems/Problem 10"
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(Cleaned up proof for Solution 1 and added an additional proof) |
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− | Thus, <math>ADC</math> and <math>ACF</math> are <math>30-60-90</math> triangles. Hence <math>AF = 15\sqrt {7}</math>, and | + | '''Key observation.''' <math>AD = 20\sqrt{7}</math>. |
− | <center><math>EF = EA + AF = 10\sqrt {7} + 15\sqrt {7} = 25\sqrt {7}</math></center> | + | |
+ | '''Proof 1.''' By the [[triangle inequality]], we can immediately see that <math>AD \geq 20\sqrt{7}</math>. However, notice that <math>10\sqrt{21} = 20\sqrt{7}\cdot\sin\frac{\pi}{3}</math>, so by the law of sines, when <math>AD = 20\sqrt{7}</math>, <math>\angle ACD</math> is right and the circle centered at <math>A</math> with radius <math>10\sqrt{21}</math>, which we will call <math>\omega</math>, is tangent to <math>\overline{CD}</math>. Thus, if <math>AD</math> were increased, <math>\overline{CD}</math> would have to be moved even farther outwards from <math>A</math> to maintain the angle of <math>\frac{\pi}{3}</math> and <math>\omega</math> could not touch it, a contradiction. | ||
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+ | '''Proof 2.''' Again, use the triangle inequality to obtain <math>AD \geq 20\sqrt{7}</math>. Let <math>x = AD</math> and <math>y = CD</math>. By the law of cosines on <math>\triangle ADC</math>, <math>2100 = x^2+y^2-xy \iff y^2-xy+(x^2-2100) = 0</math>. Viewing this as a quadratic in <math>y</math>, the discriminant <math>\Delta</math> must satisfy <math>\Delta = x^2-4(x^2-2100) = 8400-3x^2 \geq 0 \iff x \leq 20\sqrt{7}</math>. Combining these two inequalities yields the desired conclusion. | ||
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+ | This observation tells us that <math>E</math>, <math>A</math>, and <math>D</math> are collinear, in that order. | ||
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+ | Then, <math>\triangle ADC</math> and <math>\triangle ACF</math> are <math>30-60-90</math> triangles. Hence <math>AF = 15\sqrt {7}</math>, and | ||
+ | <center><math>EF = EA + AF = 10\sqrt {7} + 15\sqrt {7} = 25\sqrt {7}</math>.</center> | ||
Finally, the answer is <math>25+7=\boxed{032}</math>. | Finally, the answer is <math>25+7=\boxed{032}</math>. | ||
Revision as of 00:30, 3 January 2021
Problem
Let be an isosceles trapezoid with whose angle at the longer base is . The diagonals have length , and point is at distances and from vertices and , respectively. Let be the foot of the altitude from to . The distance can be expressed in the form , where and are positive integers and is not divisible by the square of any prime. Find .
Contents
[hide]Solution
Solution 1
Key observation. .
Proof 1. By the triangle inequality, we can immediately see that . However, notice that , so by the law of sines, when , is right and the circle centered at with radius , which we will call , is tangent to . Thus, if were increased, would have to be moved even farther outwards from to maintain the angle of and could not touch it, a contradiction.
Proof 2. Again, use the triangle inequality to obtain . Let and . By the law of cosines on , . Viewing this as a quadratic in , the discriminant must satisfy . Combining these two inequalities yields the desired conclusion.
This observation tells us that , , and are collinear, in that order.
Then, and are triangles. Hence , and
Finally, the answer is .
Solution 2
Extend through , to meet (extended through ) at . is an equilateral triangle because of the angle conditions on the base.
If then , because and therefore .
By simple angle chasing, is a 30-60-90 triangle and thus , and
Similarly is a 30-60-90 triangle and thus .
Equating and solving for , and thus .
and
See also
2008 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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.