Difference between revisions of "2013 Mock AIME I Problems/Problem 5"
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\end{align*} | \end{align*} | ||
Because <math>x>0</math>, <math>x=\sqrt{22}</math>. Thus, the perimeter of <math>ABCD</math> is <math>2x+2x+\tfrac3 2 x+x = \tfrac{13}2 x = \tfrac{13\sqrt{22}}2</math>. Thus, <math>p+q+r=13+22+2=\boxed{037}</math>. | Because <math>x>0</math>, <math>x=\sqrt{22}</math>. Thus, the perimeter of <math>ABCD</math> is <math>2x+2x+\tfrac3 2 x+x = \tfrac{13}2 x = \tfrac{13\sqrt{22}}2</math>. Thus, <math>p+q+r=13+22+2=\boxed{037}</math>. | ||
+ | |||
+ | == Solution 2 == | ||
+ | |||
+ | Consider the figure and notations from Solution 1. | ||
+ | |||
+ | Let <math>\angle DMC = \theta</math>. | ||
+ | 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> | ||
+ | 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> | ||
+ | Since <math>\cos (180^\circ - \theta) = -\cos \theta</math>, we have: | ||
+ | <cmath>\frac{25 - x^2}{24} = -\frac{80 - 4x^2}{64}.</cmath> | ||
+ | Solving for <math>x</math>, we get <math>x = \sqrt{22}</math>. | ||
+ | Now, using the cosine rule in triangles <math>AMB</math> and <math>AMD</math> with <math>\cos \theta = \frac{1}{8}</math> (substituting <math>x = \sqrt{22}</math>), we can find <math>AB</math> and <math>BC</math>. | ||
+ | After calculations, we get <math>AB = 2x</math> and <math>BC = \frac{3x}{2}</math>. | ||
+ | The perimeter of <math>ABCD</math> is given by: | ||
+ | <cmath>AB + BC + CD + DA = 2x + 2x + x + \frac{3x}{2} = \frac{13x}{2}.</cmath> | ||
+ | Substituting <math>x = \sqrt{22}</math>, we get: | ||
+ | <cmath>\text{Perimeter} = \frac{13\sqrt{22}}{2}.</cmath> | ||
+ | Therefore, <math>p = 13, q = 22, r = 2</math>, and <math>p + q + r = 13 + 22 + 2 = 37</math>. | ||
== See also == | == See also == |
Revision as of 05:57, 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,
In triangle
, by the cosine rule,
Since
, we have:
Solving for
, we get
.
Now, using the cosine rule in triangles
and
with
(substituting
), we can find
and
.
After calculations, we get
and
.
The perimeter of
is given by:
Substituting
, we get:
Therefore,
, and
.