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

Revision as of 14:41, 21 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 (No trig)

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

defaultpen(linewidth(0.6)+fontsize(11));
size(8cm);
pair A,B,C,D,P,Q,O;
A=(0,0);
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,E);
draw((10.5,15)--(10.5,0));
draw(D--(24,15),dashed);
draw(C--(30,0),dashed);
draw(D--(30,0));

label("$x$", (4.5,11.25), W);
label("$6̈$", (5.25,0), S);
label("$20$", (20.25,15), N);




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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

Video Solution

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

~Steven Chen (www.professorchenedu.com)

Video Solution 2 (Mathematical Dexterity)

https://www.youtube.com/watch?v=1nDKQkr9NaU

2022 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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