2018 AIME I Problems/Problem 15
Contents
[hide]Problem 15
David found four sticks of different lengths that can be used to form three non-congruent convex cyclic quadrilaterals, , which can each be inscribed in a circle with radius . Let denote the measure of the acute angle made by the diagonals of quadrilateral , and define and similarly. Suppose that , , and . All three quadrilaterals have the same area , which can be written in the form , where and are relatively prime positive integers. Find .
Solution 1
Suppose our four sides lengths cut out arc lengths of , , , and , where . Then, we only have to consider which arc is opposite . These are our three cases, so Our first case involves quadrilateral with , , , and .
Then, by Law of Sines, and . Therefore,
so our answer is .
Note that the conditions of the problem are satisfied when the lengths of the four sticks are about .
By S.B.
Note
The solution uses
We can see that this follows because where and are the central angles of opposite sides.
Solution 2
Suppose the four side lengths of the quadrilateral cut out arc lengths of , , , and . . Therefore, without losing generality,
, , and yields
Because Therefore,
Using the sum-to-product identities, our area of the quadrilateral then would be
Therefore, our answer is .
~Solution by eric-z
Solution 3
Let the four stick lengths be , , , and . WLOG, let’s say that quadrilateral has sides and opposite each other, quadrilateral has sides and opposite each other, and quadrilateral has sides and opposite each other. The area of a convex quadrilateral can be written as , where and are the lengths of the diagonals of the quadrilateral and is the angle formed by the intersection of and . By Ptolemy's theorem for quadrilateral , so, defining as the area of , Similarly, for quadrilaterals and , and Multiplying the three equations and rearranging, we see that The circumradius of a cyclic quadrilateral with side lengths , , , and and area can be computed as . Inserting what we know, So our answer is .
~Solution by divij04
Solution 4 (No words)
vladimir.shelomovskii@gmail.com, vvsss
Solution 5
Let the sides of the quadrilaterals be and in some order such that has opposite of , has opposite of , and has opposite of . Then, let the diagonals of be and . Similarly to solution , we get that , but this is also equal to using the area formula for a triangle using the circumradius and the sides, so and . Solving for and , we get that and , but , similarly to solution , so and the answer is .
Video Solution by MOP 2024
~r00tsOfUnity
See Also
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