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Difference between revisions of "2013 Mock AIME I Problems/Problem 5"

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<math>\boxed{037}</math>.
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Let <math>CD=x</math>, as in the diagram. Thus, from the problem, <math>BC=2x</math>. Because <math>AM \cdot MC = DM \cdot MB = 24</math>, by [[Power of a Point]], we know that <math>ABCD</math> is [[cyclic quadrilateral|cyclic]]. Thus, we know that <math>\measuredangle DAC = \measuredangle DBC</math>, so, by the congruency of vertical angles and subsequently [[AA Similarity]], we know that <math>\triangle AMD \sim \triangle BMC</math>. Thus, we have the proportion <math>\tfrac{AM}{AD} = \tfrac{BM}{BC}</math>, or, by substitution, <math>\tfrac6{AD}=\tfrac8{2x}</math>. Solving this equation for <math>AD</math> yields <math>AD=\tfrac3 2 x</math>. Similarly, we know that <math>\measuredangle ABD = \measuredangle ACD</math>, so, like before, we can see that <math>\triangle AMB \sim \triangle DMC</math>. Thus, we have the proportion <math>\tfrac{AM}{AB} = \tfrac{DM}{DC}</math>, or, by substitution, <math>\tfrac6{AB} = \tfrac3 x</math>. Solving for <math>AB</math> yields <math>AB=2x</math>.
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Now, we can use [[Ptolemy's Theorem]] on cyclic <math>ABCD</math> and solve for <math>x</math>:
 +
\begin{align*}
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x \cdot 2x + 2x \cdot \frac3 2 x &= (6+4)(8+3) \\
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5x^2 &= 110 \\
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x^2 &= 22 \\
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x &= \pm \sqrt{22}
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\end{align*}
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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>.
  
 
== See also ==
 
== See also ==

Revision as of 10:34, 30 July 2024

Problem

In quadrilateral $ABCD$, $AC\cap BD=M$. Also, $MA=6, MB=8, MC=4, MD=3$, and $BC=2CD$. The perimeter of $ABCD$ can be expressed in the form $\frac{p\sqrt{q}}{r}$ where $p$ and $r$ are relatively prime, and $q$ is not divisible by the square of any prime number. Find $p+q+r$.

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]

Let $CD=x$, as in the diagram. Thus, from the problem, $BC=2x$. Because $AM \cdot MC = DM \cdot MB = 24$, by Power of a Point, we know that $ABCD$ is cyclic. Thus, we know that $\measuredangle DAC = \measuredangle DBC$, so, by the congruency of vertical angles and subsequently AA Similarity, we know that $\triangle AMD \sim \triangle BMC$. Thus, we have the proportion $\tfrac{AM}{AD} = \tfrac{BM}{BC}$, or, by substitution, $\tfrac6{AD}=\tfrac8{2x}$. Solving this equation for $AD$ yields $AD=\tfrac3 2 x$. Similarly, we know that $\measuredangle ABD = \measuredangle ACD$, so, like before, we can see that $\triangle AMB \sim \triangle DMC$. Thus, we have the proportion $\tfrac{AM}{AB} = \tfrac{DM}{DC}$, or, by substitution, $\tfrac6{AB} = \tfrac3 x$. Solving for $AB$ yields $AB=2x$.

Now, we can use Ptolemy's Theorem on cyclic $ABCD$ and solve for $x$: \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 $x>0$, $x=\sqrt{22}$. Thus, the perimeter of $ABCD$ is $2x+2x+\tfrac3 2 x+x = \tfrac{13}2 x = \tfrac{13\sqrt{22}}2$. Thus, $p+q+r=13+22+2=\boxed{037}$.

See also

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