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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>.
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== Solution 2 ==
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Consider the figure and notations from Solution 1.
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Let <math>\angle DMC = \theta</math>.
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In triangle <math>DMC</math>, by the cosine rule,
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<cmath>\cos \theta = \frac{3^2 + 4^2 - x^2}{24} = \frac{25 - x^2}{24}.</cmath>
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In triangle <math>CMB</math>, by the cosine rule,
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<cmath>\cos (180^\circ - \theta) = \frac{4^2 + 8^2 - (2x)^2}{64} = \frac{80 - 4x^2}{64}.</cmath>
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Since <math>\cos (180^\circ - \theta) = -\cos \theta</math>, we have:
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<cmath>\frac{25 - x^2}{24} = -\frac{80 - 4x^2}{64}.</cmath>
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Solving for <math>x</math>, we get <math>x = \sqrt{22}</math>.
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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>.
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After calculations, we get <math>AB = 2x</math> and <math>BC = \frac{3x}{2}</math>.
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The perimeter of <math>ABCD</math> is given by:
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<cmath>AB + BC + CD + DA = 2x + 2x + x + \frac{3x}{2} = \frac{13x}{2}.</cmath>
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Substituting <math>x = \sqrt{22}</math>, we get:
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<cmath>\text{Perimeter} = \frac{13\sqrt{22}}{2}.</cmath>
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Therefore, <math>p = 13, q = 22, r = 2</math>, and $p + q + r = 13 + 22 + 2 = 37.
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Solution by Thunder Cloak
  
 
== See also ==
 
== See also ==
 
* [[2013 Mock AIME I Problems]]
 
* [[2013 Mock AIME I Problems]]
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* [[2013 Mock AIME I Problems/Problem 3|Preceded by Problem 4]]
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* [[2013 Mock AIME I Problems/Problem 4|Preceded by Problem 4]]
−
* [[2013 Mock AIME I Problems/Problem 5|Followed by Problem 6]]
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* [[2013 Mock AIME I Problems/Problem 6|Followed by Problem 6]]
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]

Latest revision as of 05:04, 25 August 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, $p+q+r=13+22+2=\boxed{037}$.

Solution 2

Consider the figure and notations from Solution 1.

Let $\angle DMC = \theta$. In triangle $DMC$, by the cosine rule, \[\cos \theta = \frac{3^2 + 4^2 - x^2}{24} = \frac{25 - x^2}{24}.\] In triangle $CMB$, by the cosine rule, \[\cos (180^\circ - \theta) = \frac{4^2 + 8^2 - (2x)^2}{64} = \frac{80 - 4x^2}{64}.\] Since $\cos (180^\circ - \theta) = -\cos \theta$, we have: \[\frac{25 - x^2}{24} = -\frac{80 - 4x^2}{64}.\] Solving for $x$, we get $x = \sqrt{22}$. Now, using the cosine rule in triangles $AMB$ and $AMD$ with $\cos \theta = \frac{1}{8}$ (substituting $x = \sqrt{22}$), we can find $AB$ and $BC$. After calculations, we get $AB = 2x$ and $BC = \frac{3x}{2}$. The perimeter of $ABCD$ is given by: \[AB + BC + CD + DA = 2x + 2x + x + \frac{3x}{2} = \frac{13x}{2}.\] Substituting $x = \sqrt{22}$, we get: \[\text{Perimeter} = \frac{13\sqrt{22}}{2}.\] Therefore, $p = 13, q = 22, r = 2$, and $p + q + r = 13 + 22 + 2 = 37. Solution by Thunder Cloak

See also

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