Difference between revisions of "2013 Mock AIME I Problems/Problem 5"
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== See also == | == See also == | ||
* [[2013 Mock AIME I Problems]] | * [[2013 Mock AIME I Problems]] | ||
− | * [[2013 Mock AIME I Problems/Problem | + | * [[2013 Mock AIME I Problems/Problem 4|Preceded by Problem 4]] |
− | * [[2013 Mock AIME I Problems/Problem | + | * [[2013 Mock AIME I Problems/Problem 6|Followed by Problem 6]] |
[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] |
Revision as of 11:51, 30 July 2024
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,
.