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Difference between revisions of "2022 AIME I Problems/Problem 11"

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</asy>
 
</asy>
  
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==Solution==
+
==Solution 1==
  
 
Let the circle tangent to <math>BC,AD,AB</math> at <math>P,Q,M</math> separately, denote that <math>\angle{ABC}=\angle{D}=\alpha</math>
 
Let the circle tangent to <math>BC,AD,AB</math> at <math>P,Q,M</math> separately, denote that <math>\angle{ABC}=\angle{D}=\alpha</math>
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~bluesoul
 
~bluesoul
 +
 +
==Solution 2==
 +
Denote by <math>O</math> the center of the circle. Denote by <math>r</math> the radius of the circle.
 +
Denote by <math>E</math>, <math>F</math>, <math>G</math> the points that the circle meets <math>AB</math>, <math>CD</math>, <math>AD</math> at, respectively.
 +
 +
Because the circle is tangent to <math>AD</math>, <math>CB</math>, <math>AB</math>, <math>OE = OF = OG = r</math>, <math>OE \perp AD</math>, <math>OF \perp CB</math>, <math>OG \perp AB</math>.
 +
 +
Because <math>AD \parallel CB</math>, <math>E</math>, <math>O</math>, <math>F</math> are collinear.
 +
 +
Following from the power of a point, <math>AG^2 = AE^2 = AP \cdot AQ</math>. Hence, <math>AG = AE = 6</math>.
 +
 +
Following from the power of a point, <math>CF^2 = CQ \cdot CP</math>. Hence, <math>CF = 20</math>.
 +
 +
Denote <math>BG = x</math>. Because <math>DG</math> and <math>DF</math> are tangents to the circle, <math>BF = x</math>.
 +
 +
Because <math>AEFB</math> is a right trapezoid, <math>AB^2 = EF^2 + \left( AE - BF \right)^2</math>.
 +
Hence, <math>\left( 6 + x \right)^2 = 4 r^2 + \left( 6 - x \right)^2</math>.
 +
This can be simplified as
 +
\[
 +
6 x = r^2 . \hspace{1cm} (1)
 +
\]
 +
 +
In <math>\triangle ACB</math>, by applying the law of cosines, we have
 +
\begin{align*}
 +
AC^2 & = AB^2 + CB^2 - 2 AB \cdot CB \cos B \\
 +
& = AB^2 + CB^2 + 2 AB \cdot CB \cos A \\
 +
& = AB^2 + CB^2 + 2 AB \cdot CB \cdot \frac{AE - BF}{AB} \\
 +
& = AB^2 + CB^2 + 2 CB \left( AE - BF \right) \\
 +
& = \left( 6 + x \right)^2 + \left( 20 + x \right)^2 + 2 \left( 20 + x \right) \left( 6 - x \right) \\
 +
& = 24 x + 676 .
 +
\end{align*}
 +
 +
Because <math>AC = AP + PQ + QC = 28</math>, we get <math>x = \frac{9}{2}</math>.
 +
Plugging this into Equation (1), we get <math>r = 3 \sqrt{3}</math>.
 +
 +
Therefore,
 +
\begin{align*}
 +
{\rm Area} \ ABCD & = CB \cdot EF \\
 +
& = \left( 20 + x \right) \cdot 2r \\
 +
& = 147 \sqrt{3} .
 +
\end{align*}
 +
 +
Therefore, the answer is <math>147 + 3 = \boxed{\textbf{(150) }}</math>.
 +
 +
~Steven Chen (www.professorchenedu.com)
 +
 +
==Video Solution==
 +
 +
https://www.youtube.com/watch?v=FeM_xXiJj0c&t=1s
 +
 +
~Steven Chen (www.professorchenedu.com)
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2022|n=I|num-b=10|num-a=12}}
 
{{AIME box|year=2022|n=I|num-b=10|num-a=12}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 23:57, 17 February 2022

Problem

Let $ABCD$ be a parallelogram with $\angle BAD < 90^{\circ}$. A circle tangent to sides $\overline{DA}$, $\overline{AB}$, and $\overline{BC}$ intersects diagonal $\overline{AC}$ at points $P$ and $Q$ with $AP < AQ$, as shown. Suppose that $AP = 3$, $PQ = 9$, and $QC = 16$. Then the area of $ABCD$ can be expressed in the form $m\sqrt n$, where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n$.

[asy] defaultpen(linewidth(0.6)+fontsize(11)); size(8cm); pair A,B,C,D,P,Q; A=(0,0); label("$A$", A, SW); B=(6,15); label("$B$", B, NW); C=(30,15); label("$C$", C, NE); D=(24,0); label("$D$", D, SE); P=(5.2,2.6); label("$P$", (5.8,2.6), N); Q=(18.3,9.1); label("$Q$", (18.1,9.7), W); draw(A--B--C--D--cycle); draw(C--A); draw(Circle((10.95,7.45), 7.45)); dot(A^^B^^C^^D^^P^^Q); [/asy]

Solution 1

Let the circle tangent to $BC,AD,AB$ at $P,Q,M$ separately, denote that $\angle{ABC}=\angle{D}=\alpha$

Using POP, it is very clear that $PC=20,AQ=AM=6$, let $BM=BP=x,QD=14+x$, using LOC in $\triangle{ABP}$,$x^2+(x+6)^2-2x(x+6)\cos\alpha=36+PQ^2$, similarly, use LOC in $\triangle{DQC}$, getting that $(14+x)^2+(6+x)^2-2(6+x)(14+x)\cos\alpha=400+PQ^2$. We use the second equation to minus the first equation, getting that $28x+196-(2x+12)*14*\cos\alpha=364$, we can get $\cos\alpha=\frac{2x-12}{2x+12}$.

Now applying LOC in $\triangle{ADC}$, getting $(6+x)^2+400-2(6+x)*20*\frac{2x-12}{2x+12}=(3+9+16)^2$, solving this equation to get $x=\frac{9}{2}$, then $\cos\alpha=-\frac{1}{7}$, $\sin\alpha=\frac{4\sqrt{3}}{7}$, the area is $\frac{21}{2}*\frac{49}{2}*\frac{4\sqrt{3}}{7}=147\sqrt{3}$ leads to $\boxed{150}$

~bluesoul

Solution 2

Denote by $O$ the center of the circle. Denote by $r$ the radius of the circle. Denote by $E$, $F$, $G$ the points that the circle meets $AB$, $CD$, $AD$ at, respectively.

Because the circle is tangent to $AD$, $CB$, $AB$, $OE = OF = OG = r$, $OE \perp AD$, $OF \perp CB$, $OG \perp AB$.

Because $AD \parallel CB$, $E$, $O$, $F$ are collinear.

Following from the power of a point, $AG^2 = AE^2 = AP \cdot AQ$. Hence, $AG = AE = 6$.

Following from the power of a point, $CF^2 = CQ \cdot CP$. Hence, $CF = 20$.

Denote $BG = x$. Because $DG$ and $DF$ are tangents to the circle, $BF = x$.

Because $AEFB$ is a right trapezoid, $AB^2 = EF^2 + \left( AE - BF \right)^2$. Hence, $\left( 6 + x \right)^2 = 4 r^2 + \left( 6 - x \right)^2$. This can be simplified as \[ 6 x = r^2 . \hspace{1cm} (1) \]

In $\triangle ACB$, by applying the law of cosines, we have \begin{align*} AC^2 & = AB^2 + CB^2 - 2 AB \cdot CB \cos B \\ & = AB^2 + CB^2 + 2 AB \cdot CB \cos A \\ & = AB^2 + CB^2 + 2 AB \cdot CB \cdot \frac{AE - BF}{AB} \\ & = AB^2 + CB^2 + 2 CB \left( AE - BF \right) \\ & = \left( 6 + x \right)^2 + \left( 20 + x \right)^2 + 2 \left( 20 + x \right) \left( 6 - x \right) \\ & = 24 x + 676 . \end{align*}

Because $AC = AP + PQ + QC = 28$, we get $x = \frac{9}{2}$. Plugging this into Equation (1), we get $r = 3 \sqrt{3}$.

Therefore, \begin{align*} {\rm Area} \ ABCD & = CB \cdot EF \\ & = \left( 20 + x \right) \cdot 2r \\ & = 147 \sqrt{3} . \end{align*}

Therefore, the answer is $147 + 3 = \boxed{\textbf{(150) }}$.

~Steven Chen (www.professorchenedu.com)

Video Solution

https://www.youtube.com/watch?v=FeM_xXiJj0c&t=1s

~Steven Chen (www.professorchenedu.com)

See Also

2022 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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