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).
Solution 7 Analytic Geometry
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 .
Since and , and
With law of cosines,
Since , is acute angle. and
Let line be axis.
Line equation is .
Since line passes point and perpendicular to , its equation is
where ,
Since is the intersection of and ,
(Negative means point is between point and )
and the answer is .
Note: if was longer, point would be between point and . Then, would be the diagonal of quadrilateral not the side. To apply Ptolemy's theorem like solution 4, it is critical to know whether is the diagonal or side of quadrilateral. Equation for wrong case cannot be solved. For example,
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 | ||
All AIME Problems and Solutions |
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