Difference between revisions of "2021 AIME II Problems/Problem 12"
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</cmath> | </cmath> | ||
− | Because <math>\cos \angle AEB = \cos \left( 180^\circ - \theta \right) = | + | Because <math>\cos \angle AEB = \cos \left( 180^\circ - \theta \right) = \cos \theta</math> and <math>AB = 5</math>, we have |
<cmath> | <cmath> | ||
\[ | \[ | ||
Line 50: | Line 50: | ||
</cmath> | </cmath> | ||
− | + | In <math>\triangle BEC</math>, following from the law of cosines, we have | |
− | + | <cmath> | |
+ | \[ | ||
+ | b^2 + c^2 - 2 b c \cos \angle BEC | ||
+ | = BC^2 . | ||
+ | \] | ||
+ | </cmath> | ||
+ | |||
+ | Because <math>\cos \angle AEB = \cos \theta</math> and <math>BC = 6</math>, we have | ||
+ | <cmath> | ||
+ | \[ | ||
+ | b^2 + c^2 - 2 b c \cos \theta = 6^2 . \ \ \ (3) | ||
+ | \] | ||
+ | </cmath> | ||
+ | |||
+ | In <math>\triangle CED</math>, following from the law of cosines, we have | ||
+ | <cmath> | ||
+ | \[ | ||
+ | c^2 + d^2 - 2 c d \cos \angle CED | ||
+ | = CD^2 . | ||
+ | \] | ||
+ | </cmath> | ||
− | + | Because <math>\cos \angle CED = \cos \left( 180^\circ - \theta \right) = \cos \theta</math> and <math>CD = 9</math>, we have | |
+ | <cmath> | ||
+ | \[ | ||
+ | c^2 + d^2 + 2 c d \cos \theta = 9^2 . \ \ \ (4) | ||
+ | \] | ||
+ | </cmath> | ||
− | + | In <math>\triangle DEA</math>, following from the law of cosines, we have | |
+ | <cmath> | ||
+ | \[ | ||
+ | d^2 + a^2 - 2 d a \cos \angle DEA | ||
+ | = DA^2 . | ||
+ | \] | ||
+ | </cmath> | ||
− | + | Because <math>\cos \angle DEC = \cos \theta</math> and <math>DA = 7</math>, we have | |
+ | <cmath> | ||
+ | \[ | ||
+ | d^2 + a^2 - 2 d a \cos \theta = 7^2 . \ \ \ (5) | ||
+ | \] | ||
+ | </cmath> | ||
By taking Eq (2) - Eq (3) + Eq (4) - Eq (5) and dividing both sides of the equation by 2, we get <math>\left( ab + bc + cd + da \right) \cos \theta = \frac{21}{2}</math>. We index this equation as Eq (6). | By taking Eq (2) - Eq (3) + Eq (4) - Eq (5) and dividing both sides of the equation by 2, we get <math>\left( ab + bc + cd + da \right) \cos \theta = \frac{21}{2}</math>. We index this equation as Eq (6). |
Revision as of 19:44, 22 March 2021
Contents
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 where the second equality follows from the formula to use the sine function to compute a triangle area, the the fourth equality follows from the property that .
Because , we have .
Second, we use the law of cosines to establish four equations for four sides of the quadrilateral .
In , following from the law of cosines, we have
Because and , we have
In , following from the law of cosines, we have
Because and , we have
In , following from the law of cosines, we have
Because and , we have
In , following from the law of cosines, we have
Because and , we have
By taking Eq (2) - Eq (3) + Eq (4) - Eq (5) and dividing both sides of the equation by 2, we get . We index this equation as Eq (6).
By taking , we get .
Therefore, by writing this answer in the form of , we have and . Therefore, the answer to this question is .
~ Steven Chen (www.professorchenedu.com)
Solution 2
Since we are asked to find , we can find and separately and use their values to get . We can start by drawing a diagram. Let the vertices of the quadrilateral be , , , and . Let , , , and . Let , , , and . We know that is the acute angle formed between the intersection of the diagonals and .
We are given that the area of quadrilateral is . We can express this area using the areas of triangles , , , and . Since we want to find and , we can represent these areas using as follows:
We know that . Therefore it follows that:
From here we see that . Now we need to find . Using the Law of Cosines on each of the four smaller triangles, we get following equations:
We know that . We can substitute this value into our equations to get:
If we subtract the sum of the first and third equation from the sum of the second and fourth equation, the squared terms cancel, leaving us with:
From here we see that .
Since we have figured out and , we can calculate :
Therefore our answer is .
~ my_aops_lessons
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.