Difference between revisions of "2018 AIME I Problems/Problem 15"
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==Problem 15== | ==Problem 15== | ||
− | David found four sticks of different lengths that can be used to form three non-congruent convex cyclic quadrilaterals, <math>A,\text{ }B,\text{ }C</math>, which can each be inscribed in a circle with radius <math>1</math>. Let <math>\varphi_A</math> denote the measure of the acute angle made by the diagonals of quadrilateral <math>A</math>, and define <math>\varphi_B</math> and <math>\varphi_C</math> similarly. Suppose that <math>\sin\varphi_A=\ | + | David found four sticks of different lengths that can be used to form three non-congruent convex cyclic quadrilaterals, <math>A,\text{ }B,\text{ }C</math>, which can each be inscribed in a circle with radius <math>1</math>. Let <math>\varphi_A</math> denote the measure of the acute angle made by the diagonals of quadrilateral <math>A</math>, and define <math>\varphi_B</math> and <math>\varphi_C</math> similarly. Suppose that <math>\sin\varphi_A=\tfrac{2}{3}</math>, <math>\sin\varphi_B=\tfrac{3}{5}</math>, and <math>\sin\varphi_C=\tfrac{6}{7}</math>. All three quadrilaterals have the same area <math>K</math>, which can be written in the form <math>\dfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>. |
==Solution 1== | ==Solution 1== |
Revision as of 15:49, 26 December 2022
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.
Solution 2
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 3 (No words)
vladimir.shelomovskii@gmail.com, vvsss
Solution 4
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 .
See Also
2018 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last question | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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