Difference between revisions of "2013 Mock AIME I Problems/Problem 5"
Thundercloak (talk | contribs) (→Solution 2) |
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Let <math>\angle DMC = \theta</math>. | Let <math>\angle DMC = \theta</math>. | ||
In triangle <math>DMC</math>, by the cosine rule, | In triangle <math>DMC</math>, by the cosine rule, | ||
− | <cmath>\cos \theta = \frac{3^2 + 4^2 - x^2}{24} = \frac{25 - x^2}{24}.</cmath>. | + | <cmath>\cos \theta = \frac{3^2 + 4^2 - x^2}{24} = \frac{25 - x^2}{24}.</cmath>. (1) |
In triangle <math>CMB</math>, by the cosine rule, | In triangle <math>CMB</math>, by the cosine rule, | ||
<cmath>\cos (180^\circ - \theta) = \frac{4^2 + 8^2 - (2x)^2}{64} = \frac{80 - 4x^2}{64}.</cmath> | <cmath>\cos (180^\circ - \theta) = \frac{4^2 + 8^2 - (2x)^2}{64} = \frac{80 - 4x^2}{64}.</cmath> | ||
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Since <math>\cos (180^\circ - \theta) = -\cos \theta</math>, we have: | Since <math>\cos (180^\circ - \theta) = -\cos \theta</math>, we have: | ||
<cmath>\frac{25 - x^2}{24} = -\frac{80 - 4x^2}{64}.</cmath> | <cmath>\frac{25 - x^2}{24} = -\frac{80 - 4x^2}{64}.</cmath> |
Revision as of 05:01, 25 August 2024
Contents
Problem
In quadrilateral , . Also, , and . The perimeter of can be expressed in the form where and are relatively prime, and is not divisible by the square of any prime number. Find .
Solution
Let , as in the diagram. Thus, from the problem, . Because , by Power of a Point, we know that is cyclic. Thus, we know that , so, by the congruency of vertical angles and subsequently AA Similarity, we know that . Thus, we have the proportion , or, by substitution, . Solving this equation for yields . Similarly, we know that , so, like before, we can see that . Thus, we have the proportion , or, by substitution, . Solving for yields .
Now, we can use Ptolemy's Theorem on cyclic and solve for : \begin{align*} x \cdot 2x + 2x \cdot \frac3 2 x &= (6+4)(8+3) \\ 5x^2 &= 110 \\ x^2 &= 22 \\ x &= \pm \sqrt{22} \end{align*} Because , . Thus, the perimeter of is . Thus, .
Solution 2
Consider the figure and notations from Solution 1.
Let . In triangle , by the cosine rule, . (1) In triangle , by the cosine rule, Since , we have: Solving for , we get . Now, using the cosine rule in triangles and with (substituting in (1)), we can find and . After calculations, we get and . The perimeter of is given by: Substituting , we get: Therefore, , and .