# 2022 AIME I Problems/Problem 11

## 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 (No trig)

Let's redraw the diagram, but extend some helpful lines.

$[asy] size(20cm); pair A,B,C,D,E,F,P,Q,O; A=(0,0); E = (24,15); F = (30,0); O = (10.5,7.5); label("A", A, SW); B=(6,15); label("B", B, NW); C=(30,15); label("C", C, NE); D=(24,0); label("D", D, SE); P=(5.2,2.6); label("P", (5.8,2.6), N); Q=(18.3,9.1); label("Q", (18.1,9.7), W); draw(A--B--C--D--cycle); draw(C--A); draw(Circle((10.95,7.45), 7.45)); dot(A^^B^^C^^D^^P^^Q); dot(O); label("O",O,W); draw((10.5,15)--(10.5,0)); draw(D--(24,15),dashed); draw(C--(30,0),dashed); draw(D--(30,0)); dot(E); dot(F); label("3", midpoint(A--P), S); label("9", midpoint(P--Q), S); label("16", midpoint(Q--C), S); label("x", (5.5,13.75), W); label("20", (20.25,15), N); label("6", (5.25,0), S); label("6", (1.5,3.75), W); label("x", (8.25,15),N); label("14+x", (17.25,0), S); label("6-x", (27,15), N); label("6+x", (27,7.5), W); label("6\sqrt{3}", (30,7.5),W); label("T_1", (10.5,15), N); label("T_2", (10.5,0), S); label("T_3", (4.5,11.25),W); label("E",E, N); label("F",F, S); [/asy]$

We obviously see that we must use power of a point since they've given us lengths in a circle and there are intersection points. Let $T_1, T_2, T_3$ be our tangents from the circle to the parallelogram. By the secant power of a point, the power of $A = 3 \cdot (3+9) = 36$. Then $AT_2 = AT_3 = \sqrt{36} = 6$. Similarly, the power of $C = 16 \cdot (16+9) = 400$ and $CT_1 = \sqrt{400} = 20$. We let $BT_3 = BT_1 = x$ and label the diagram accordingly.

Notice that because $BC = AD, 20+x = 6+DT_2 \implies DT_2 = 14+x$. Let $O$ be the center of the circle. Since $OT_1$ and $OT_2$ intersect $BC$ and $AD$, respectively, at right angles, we have $T_2T_2CD$ is a right-angled trapezoid and more importantly, the diameter of the circle is the height of the triangle. Therefore, we can drop an altitude from $D$ to $BC$ and $C$ to $AD$, and both are equal to $2r$. Since $T_1E = T_2D$, $20 - CE = 14+x \implies CE = 6-x$. Since $CE = DF, DF = 6-x$ and $AF = 6+14+x+6-x = 26$. We can now use Pythagorean theorem on $\triangle ACF$; we have $26^2 + (2r)^2 = (3+9+16)^2 \implies 4r^2 = 784-676 \implies 4r^2 = 108 \implies 2r = 6\sqrt{3}$ and $r^2 = 27$.

We know that $CD = 6+x$ because $ABCD$ is a parallelogram. Using Pythagorean theorem on $\triangle CDF$, $(6+x)^2 = (6-x)^2 + 108 \implies (6+x)^2-(6-x)^2 = 108 \implies 12 \cdot 2x = 108 \implies 2x = 9 \implies x = \frac{9}{2}$. Therefore, base $BC = 20 + \frac{9}{2} = \frac{49}{2}$. Thus the area of the parallelogram is the base times the height, which is $\frac{49}{2} \cdot 6\sqrt{3} = 147\sqrt{3}$ and the answer is $\boxed{150}$

~KingRavi

## Solution 2

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 3

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)

## Solution 4

Let $\omega$ be the circle, let $r$ be the radius of $\omega$, and let the points at which $\omega$ is tangent to $AB$, $BC$, and $AD$ be $X$, $Y$, and $Z$, respectively. Note that PoP on $A$ and $C$ with respect to $\omega$ yields $AX=6$ and $CY=20$. We can compute the area of $ABC$ in two ways:

1. By the half-base-height formula, $[ABC]=r(20+BX)$.

2. We can drop altitudes from the center $O$ of $\omega$ to $AB$, $BC$, and $AC$, which have lengths $r$, $r$, and $\sqrt{r^2-81/4}$. Thus, $[ABC]=[OAB]+[OBC]+[OAC]=r(BX+13)+14\sqrt{r^2-81/4}$.

Equating the two expressions for $[ABC]$ and solving for $r$ yields $r=3\sqrt{3}$.

Let $BX=BY=a$. By the Parallelogram Law, $(a+6)^2+(a+20)^2=38^2$. Solving for $a$ yields $a=9/2$. Thus, $[ABCD]=2[ABC]=2r(20+a)=147\sqrt{3}$, for a final answer of $\boxed{150}$.

~ Leo.Euler

## Solution 5

Let $\omega$ be the circle, let $r$ be the radius of $\omega$, and let the points at which $\omega$ is tangent to $AB$, $BC$, and $AD$ be $H$, $K$, and $T$, respectively. PoP on $A$ and $C$ with respect to $\omega$ yields $$AT=6, CK=20.$$

Let $TG = AC, CG||AT.$

In $\triangle KGT$ $KT \perp BC,$ $KT = \sqrt{GT^2 – (KC + AT)^2} = 6 \sqrt{3}=2r.$

$\angle AOB = 90^{\circ}, OH \perp AB, OH = r = \frac{KT}{2},$ $$OH^2 = AH \cdot BH \implies BH = \frac {9}{2}.$$

Area is $$(BK + KC) \cdot KT = (BH + KC) \cdot 2r = \frac{49}{2} \cdot 6\sqrt{3} = 147 \sqrt{3} \implies 147+3 = \boxed{\textbf{150}}.$$

## Video Solution

~Steven Chen (www.professorchenedu.com)