Difference between revisions of "2021 AIME II Problems/Problem 12"
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First, we write down an equation of the area of the quadrilateral <math>ABCD</math>. | First, we write down an equation of the area of the quadrilateral <math>ABCD</math>. | ||
− | We have <math>{\rm Area} \ ABCD = {\rm Area} \ \triangle | + | We have <math>{\rm Area} \ ABCD = {\rm Area} \ \triangle ABE + {\rm Area} \ \triangle BCE + {\rm Area} \ \triangle CDE + {\rm Area} \ \triangle DAE = \frac{1}{2} ab \sin \angle AEB + \frac{1}{2} bc \sin \angle BEC + \frac{1}{2} cd \sin \angle CED + \frac{1}{2} da \sin \angle DEA = \frac{1}{2} \left( ab + bc + cd + da \right) \sin \theta</math>. |
+ | Because <math>{\rm Area} \ ABCD = 30</math>, we have <math>\left( ab + bc + cd + da \right) \sin \theta = 60</math>. We index this equation as Eq (1). | ||
==See also== | ==See also== | ||
{{AIME box|year=2021|n=II|num-b=11|num-a=13}} | {{AIME box|year=2021|n=II|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 17:39, 22 March 2021
Problem
A convex quadrilateral has area and side lengths and in that order. Denote by the measure of the acute angle formed by the diagonals of the quadrilateral. Then can be written in the form , where and are relatively prime positive integers. Find .
Solution 1
We denote by , , and four vertices of this quadrilateral, such that , , , . We denote by the point that two diagonals and meet at. To simplify the notation, we denote , , , . We denote .
First, we write down an equation of the area of the quadrilateral . We have . Because , we have . We index this equation as Eq (1).
See also
2021 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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