2013 Mock AIME I Problems/Problem 5
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
![[asy] import geometry; // Defining Points point O = origin; point B = (1,0); point A = dir(115.583); point C = dir(-115.583); point D = dir(-165.638); point M; // Circle draw(circle(O, 1)); // Quadrilateral and Diagonals draw(A--B--C--D--cycle); draw(A--C); draw(B--D); // Defining M pair[] m = intersectionpoints((A--C),(B--D)); M = m[0]; // Labelling Points dot(A); label("A",A,NW); dot(B); label("B",B,E); dot(C); label("C",C,SW); dot(D); label("D",D,WSW); dot(M); label("M",M,NE); // Length Labels label("$3$", midpoint(D--M), NNW); label("$8$", midpoint(M--B), NNW); label("$6$", midpoint(A--M), E); label("$4$", midpoint(C--M), E); label("$2x$", midpoint(B--C), SE); label("$x$", midpoint(C--D), NE); [/asy]](http://latex.artofproblemsolving.com/e/9/f/e9fbbaf2edd2f0b3d51e0accf2c6b0465cc8b226.png)
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, 
.
