2011 AIME II Problems/Problem 10
Contents
Problem 10
A circle with center has radius 25. Chord of length 30 and chord of length 14 intersect at point . The distance between the midpoints of the two chords is 12. The quantity can be represented as , where and are relatively prime positive integers. Find the remainder when is divided by 1000.
Solution 1
Let and be the midpoints of and , respectively, such that intersects .
Since and are midpoints, and .
and are located on the circumference of the circle, so .
The line through the midpoint of a chord of a circle and the center of that circle is perpendicular to that chord, so and are right triangles (with and being the right angles). By the Pythagorean Theorem, , and .
Let , , and be lengths , , and , respectively. OEP and OFP are also right triangles, so , and
We are given that has length 12, so, using the Law of Cosines with :
Substituting for and , and applying the Cosine of Sum formula:
and are acute angles in right triangles, so substitute opposite/hypotenuse for sines and adjacent/hypotenuse for cosines:
Combine terms and multiply both sides by :
Combine terms again, and divide both sides by 64:
Square both sides:
This reduces to ; .
Solution 2 - Fastest
We begin as in the first solution. Once we see that has side lengths , , and , we can compute its area with Heron's formula:
Thus, the circumradius of triangle is . Looking at , we see that , which makes it a cyclic quadrilateral. This means 's circumcircle and 's inscribed circle are the same.
Since is cyclic with diameter , we have , so and the answer is .
Solution 3
We begin as the first solution have and . Because , Quadrilateral is inscribed in a Circle. Assume point is the center of this circle.
point is on
Link and , Made line , then
On the other hand,
As a result,
Therefore,
As a result,
Solution 4
Let .
Proceed as the first solution in finding that quadrilateral has side lengths , , , and , and diagonals and .
We note that quadrilateral is cyclic and use Ptolemy's theorem to solve for :
Solving, we have so the answer is .
-Solution by blueberrieejam
~bluesoul changes the equation to a right equation, the previous equation isn't solvable
Solution 5 (Quick Angle Solution)
Let be the midpoint of and of . As , quadrilateral is cyclic with diameter . By Cyclic quadrilaterals note that .
The area of can be computed by Herons as The area is also . Therefore,
~ Aaryabhatta1
Solution 6
Define and as the midpoints of and , respectively. Because , we have that is a cyclic quadrilateral. Hence, Then, let these two angles be denoted as . Now, assume WLOG that and (We can do this because one of or must be less than 7, and similarly for and ). Then, by Power of a Point on P with respect to the circle with center , we have that Then, let . From Law of Cosines on , we have that Plugging in in gives Hence, Then, we also know that Squaring this, we get Equating our expressions for , we get Solving gives us that . Since , from the Pythagorean Theorem, , and thus the answer is , which when divided by a thousand leaves a remainder of
-Mr.Sharkman
Note: my solution was very long and tedious. It was definitely was the least elegant solution. The only thing I like about it is it contains no quadratic equations (unless you count LoC).
See also
2011 AIME II (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 | ||
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