Difference between revisions of "2013 Mock AIME I Problems/Problem 5"
Thundercloak (talk | contribs) (→Solution) |
Thundercloak (talk | contribs) (→Solution 2) |
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Substituting <math>x = \sqrt{22}</math>, we get: | Substituting <math>x = \sqrt{22}</math>, we get: | ||
<cmath>\text{Perimeter} = \frac{13\sqrt{22}}{2}.</cmath> | <cmath>\text{Perimeter} = \frac{13\sqrt{22}}{2}.</cmath> | ||
− | Therefore, <math>p = 13, q = 22, r = 2</math>, and | + | Therefore, <math>p = 13, q = 22, r = 2</math>, and $p + q + r = 13 + 22 + 2 = 37. |
+ | Solution by Thunder Cloak | ||
== See also == | == See also == |
Latest revision as of 06:04, 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 $p + q + r = 13 + 22 + 2 = 37.
Solution by Thunder Cloak