Difference between revisions of "2021 AIME II Problems/Problem 12"
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We denote <math>\theta = \angle AED</math>. | We denote <math>\theta = \angle AED</math>. | ||
− | First, we write down an equation of the area of the quadrilateral <math>ABCD</math>. | + | First, we use the triangle area formula with sines to write down an equation of the area of the quadrilateral <math>ABCD</math>. |
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>. | 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). | 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). | ||
+ | |||
+ | Second, we use the law of cosines to establish four equations for four sides of the quadrilateral <math>ABCD</math>. | ||
+ | |||
+ | By applying the law of cosines to <math>\triangle AEB</math>, we have <math>a^2 + b^2 - 2 a b \cos \angle AEB = AB^2 = 5^2</math>. | ||
+ | Note that <math>\cos \angle AEB = \cos \left( 180^\circ - \theta \right) = \cos \theta</math>. | ||
+ | Hence, <math>a^2 + b^2 + 2 a b \cos \theta = 5^2</math>. | ||
+ | We index this equation as Eq (2). | ||
+ | |||
+ | Analogously, we can establish the following equation for <math>\triangle BEC</math> that <math>b^2 + c^2 - 2 b c \cos \theta = 6^2</math> (indexed as Eq (3)), the following equation for <math>\triangle CED</math> that <math>c^2 + d^2 + 2 c d \cos \theta = 9^2</math> (indexed as Eq (4)) and the following equation for <math>\triangle DEA</math> that <math>d^2 + a^2 - 2 d a \cos \theta = 7^2</math>. | ||
+ | |||
+ | ~ Steven Chen (from Professor Chen Education Palace) | ||
==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 17:45, 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 use the triangle area formula with sines to write down an equation of the area of the quadrilateral . We have .
Because , we have . We index this equation as Eq (1).
Second, we use the law of cosines to establish four equations for four sides of the quadrilateral .
By applying the law of cosines to , we have . Note that . Hence, . We index this equation as Eq (2).
Analogously, we can establish the following equation for that (indexed as Eq (3)), the following equation for that (indexed as Eq (4)) and the following equation for that .
~ Steven Chen (from Professor Chen Education Palace)
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.