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

Latest revision as of 18:33, 6 January 2025

1 Problem
2 Solution 1 (No trig)
3 Solution 2
4 Solution 3
5 Solution 4
6 Solution 5
7 Solution 6 (Short and Sweet)
8 Solution 7 (Intuitive, no trig, no weird auxiliary lines)
9 Solution 8 (Ptolemy's Theorem + Power of Point + Pythagorean Theorem)
10 Solution 9
11 Video Solution
12 Video Solution 2 (Mathematical Dexterity)
13 Video Solution 3 by OmegaLearn
14 See Also

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.

$[asy] size(10cm); pair A,B,C,D,EE,F,P,Q,O; A=(0,0); EE = (24,15); F = (30,0); O = (10.5,7.5); label("$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(EE); 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), S); label("$6\sqrt{3}$", (30,7.5), E); label("$T_1$", (10.5,15), N); label("$T_2$", (10.5,0), S); label("$T_3$", (4.5,11.25), W); label("$E$", EE, 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 $T_2T_1CD$ 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 $28x+196-(2x+12)\times14\times\cos\alpha=364$ , we can get $\cos\alpha=\frac{2x-12}{2x+12}$ .

Now applying LOC in $\triangle{ADC}$ , getting $(6+x)^2+(20+x)^2-2(6+x)\times(20+x)\times\frac{2x-12}{2x+12}=(3+9+16)^2$ , 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 $\frac{21}{2}\cdot\frac{49}{2}\cdot\frac{4\sqrt{3}}{7}=147\sqrt{3}$ leads to $\boxed{150}$

~bluesoul,HarveyZhang

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 $\sqrt{r^2-\frac{81}{4}}$ . Thus, $[ABC]=[OAB]+[OBC]+[OAC]=r(BX+13)+14\sqrt{r^2-\frac{81}{4}}$ .

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

vladimir.shelomovskii@gmail.com, vvsss

Solution 6 (Short and Sweet)

Let $O$ be the center of the circle. Let points $M, N$ and $L$ be the tangent points of lines $BC, AD$ and $AB$ respectively to the circle. By Power of a Point, $({MC})^2=16\cdot{25} \Longrightarrow MC=20$ . Similarly, $({AL})^2=3\cdot{12} \Longrightarrow AL=6$ . Notice that $AL=AN=6$ since quadrilateral $LONA$ is symmetrical. Let $AC$ intersect $MN$ at $I$ . Then, $\bigtriangleup{IMC}$ is similar to $\bigtriangleup{AIN}$ . Therefore, $\frac{CI}{MC}=\frac{AI}{AN}$ . Let the length of $PI=l$ , then $\frac{25-l}{20}=\frac{3+l}{6}$ . Solving we get $l=\frac{45}{13}$ . Doing the Pythagorean theorem on triangles $IMC$ and $AIN$ for sides $MI$ and $IN$ respectively, we obtain the equation $\sqrt{(\frac{280}{13})^2-400} +\sqrt{(\frac{84}{13})^2-36}=MN=2r_1$ where $r_1$ denotes the radius of the circle. Solving, we get $MN=6\sqrt{3}$ . Additionally, quadrilateral $OLBM$ is symmetrical so $OL=OM$ . Let $OL=OM=x$ and extend a perpendicular foot from $B$ to $AD$ and call it $R$ . Then, $\bigtriangleup{ABR}$ is right with $AR=6-x$ , $AB=6+x$ , and $RB=2r_1=MN=6\sqrt{3}$ . Taking the difference of squares, we get $108=24x \Longrightarrow x=\frac{9}{2}$ . The area of $ABCD$ is $MN\cdot{BC}=(20+x)\cdot{MN} \Longrightarrow \frac{49}{2}\cdot{6\sqrt{3}}=147\sqrt{3}$ . Therefore, the answer is $147+3=\boxed{150}$

~Magnetoninja

Solution 7 (Intuitive, no trig, no weird auxiliary lines)

Say that $BC$ is tangent to the circle at $X$ and $AD$ tangent at $Y$ . Also, $H$ is the intersection of $XY$ (diameter) and $AC$ (diagonal). Then by power of a point with given info on $A$ and $C$ we get that $AY=6$ and $CX=20$ . Note that $HAY \sim HCX$ , and since $\frac{AY}{CX}=\frac{3}{10}$ we note that $\frac{AH}{CH} = \frac{AP+PH}{CQ+QH} = \frac{3+PH}{16+QH} =\frac{AY}{CX}=\frac{3}{10}$ . Since $PH+HQ=9$ , we get that $PH=\frac{45}{13}$ and $QH=\frac{72}{13}$ . This is the length information within the circle. The same triangle similarity also means that $\frac{YH}{XH}=\frac{3}{10}$ , so if the radius of the circle is $r$ then we have $XH=\frac{20}{13}r$ and $YH = \frac{6}{13}r$ . By power of a point on H, we can figure out $r$ : $XH\cdot YH = PH \cdot QG$ $\frac{20}{13}r \cdot \frac{6}{13}r = \frac{45}{13} \cdot \frac{72}{13}$ and we get that $r = 3 \sqrt 3$ . Thus, we have that the height of the parallelogram is $2r=6 \sqrt 3$ and we want to find $BC$ . If $AB$ is tangent to the circle at $E$ , then set $a = BX = BE$ . Using pythagorean theorem, $AO^2+BO^2=AB^2$ and we can plug in diagram values: $(AY^2+OY^2)+(BX^2+OX)^2=AB^2$ $(6^2+(3 \sqrt 3)^2) + (a^2+(3 \sqrt 3)^2)=(a+6)^2.$ Solving, we get $a=\frac{9}{2}$ Finally, we have $[ABCD]=XY \cdot BC = 6 \sqrt 3 \cdot (20+\frac{9}{2}) \rightarrow \boxed{150}$

~ Brocolimanx

Solution 8 (Ptolemy's Theorem + Power of Point + Pythagorean Theorem)

Let $E$ , $F$ , $G$ be the circle's point of tangency with sides $AD$ , $AB$ , and $BC$ , respectively. Let $O$ be the center of the inscribed circle.

By Power of a Point, $AE^2 = AP \cdot AQ = 3(3+9) = 36$ , so $AE = 6$ . Similarly, $GC^2 = CQ \cdot CP = 16(16+9) = 400$ , so $GC = 20$ .

Construct $GE$ , and let $I$ be the point of intersection of $GE$ and $AC$ . $GE \perp BC$ and $GE \perp AD$ . By AA, $\triangle IGC \sim \triangle IEA$ , and we have $\frac{AI}{IC} = \frac{AE}{GC} = \frac{3}{10}$ . We also know $AI + IC = AC = 28$ , so $AI = \frac{84}{13}$ and $IC = \frac{280}{13}$ .

Using Pythagorean Theorem on $\triangle IEA$ and $\triangle CIG$ , we find that $EI = \frac{18\sqrt{3}}{13}$ and $IG = \frac{60\sqrt{3}}{13}$ . Thus, $GE = EI + IG = 6\sqrt{3}$ , and the radius of the circle is $3\sqrt{3}$ .

Construct $EF$ , $FG$ . $\angle AFO = \angle AEO = 90^{\circ}$ , so $AEOF$ is cyclic. Similarly, $BFOG$ is cyclic.

Now, we attempt to set up Ptolemy. Using Pythagorean Theorem on $\triangle AEO$ , we find that $AO = 3\sqrt{7}$ . By Ptolemy's Theorem, $(AE)(FO) + (AF)(EO) = (AO)(FE)$ , from which we have $(6)(3\sqrt{3}) + (6)(3\sqrt{3}) = (3\sqrt{7})(FE)$ and $FE = 12\frac{\sqrt{3}}{\sqrt{7}}$ . From Thales' Circle, $\triangle FGE$ is a right triangle, and $EF^2 + FG^2 = GE^2$ , so $FG = \frac{18}{\sqrt{7}}$ .

Set $BF = BG = s$ . $BO = \sqrt{s^2 + (3\sqrt{3})^2} = \sqrt{s^2+27}$ , so by Ptolemy's Theorem on $BFOG$ , we have

$(BF)(GO) + (BG)(FO) = (FG)(BO)$ $(3\sqrt{3})(s) + (3\sqrt{3})(s) = (\frac{18}{\sqrt{7}})(\sqrt{s^2+27})$ Solving yields $s = \frac{9}{2}$ .

We know that $BC = BG + GC = 20 + \frac{9}{2} = \frac{49}{2}$ , so the area of $ABCD = (\frac{49}{2})(6\sqrt{3}) = 147\sqrt{3}$ . The requested answer is $147 + 3 = \boxed{150}$ .

~ adam_zheng

Solution 9

Let points $E$ , $F$ , $G$ be the points where lines $BC$ , $AB$ , and $AD$ are tangent to the circle respectively. Then extend line $BC$ to point $H$ such that $AH$ is perpendicular to $BC$ .

According to Power of a Point, $CQ\cdot CP=CE^2\rightarrow 16\cdot 25=CE^2\rightarrow CE=20$ .

Similarly, $AP\cdot AQ=AF^2\rightarrow 3\cdot 12=AF^2\rightarrow AF=6$ .

(Note that $AG=AF=6$ because they are intersecting tangents from the same circle. Furthermore, $HE=AG=6$ due to a simple upwards translation.)

Therefore, $HC=HE+EC=6+20=26$ , and we already know that $AC=28$ , meaning that we can use the Pythagorean Theorem on $\Delta AHC$ to obtain: $AH=6\sqrt{3}$ .

Let $HB=k$ . Then $BE=6-HB=6-k$ . Since they are intersecting tangents from the same circle, $BF=BE=6-k$ .

Therefore, $AB=AF+BF=6+6-k=12-k$ . We have all the side lengths of $\Delta ABH$ , so applying the Pythagorean Theorem: $(6\sqrt{3})^2+k^2=(12-k)^2\rightarrow k=\frac{3}{2}$ .

Thus, $BC=BE+EC=6-k+20=26-k=\frac{49}{2}$ .

$[ABCD]=AH\cdot BC=6\sqrt{3}\cdot \frac{49}{2}=147\sqrt{3}$ , so the answer is $\boxed{150}$ .

~sid2012

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

Video Solution 3 by OmegaLearn

https://youtu.be/LpOegT0fKy8?t=740

~ pi_is_3.14

@@ Line 1: / Line 1: @@
 == Problem ==
-Let <math>ABCD</math> be a parallelogram with <math>\angle BAD < 90^{\circ}</math>. A circle tangent to sides <math>\overline{DA}</math>, <math>\overline{AB}</math>, and <math>\overline{BC}</math> intersects diagonal <math>\overline{AC}</math> at points <math>P</math> and <math>Q</math> with <math>AP < AQ</math>, as shown. Suppose that <math>AP = 3</math>, <math>PQ = 9</math>, and <math>QC = 16</math>. Then the area of <math>ABCD</math> can be expressed in the form <math>m\sqrt n</math>, where <math>m</math> and <math>n</math> are positive integers, and <math>n</math> is not divisible by the square of any prime. Find <math>m+n</math>.
+Let <math>ABCD</math> be a parallelogram with <math>\angle BAD < 90^\circ.</math> A circle tangent to sides <math>\overline{DA},</math> <math>\overline{AB},</math> and <math>\overline{BC}</math> intersects diagonal <math>\overline{AC}</math> at points <math>P</math> and <math>Q</math> with <math>AP < AQ,</math> as shown. Suppose that <math>AP=3,</math> <math>PQ=9,</math> and <math>QC=16.</math> Then the area of <math>ABCD</math> can be expressed in the form <math>m\sqrt{n},</math> where <math>m</math> and <math>n</math> are positive integers, and <math>n</math> is not divisible by the square of any prime. Find <math>m+n.</math>
 <asy>
@@ Line 29: / Line 29: @@
 <asy>
-size(20cm);
+size(10cm);
-pair A,B,C,D,E,F,P,Q,O;
+pair A,B,C,D,EE,F,P,Q,O;
 A=(0,0);
-E = (24,15);
+EE = (24,15);
 F = (30,0);
 O = (10.5,7.5);
@@ Line 56: / Line 56: @@
 draw(C--(30,0),dashed);
 draw(D--(30,0));
-dot(E);
+dot(EE);
 dot(F);
@@ Line 69: / Line 69: @@
 label("$14+x$", (17.25,0), S);
 label("$6-x$", (27,15), N);
-label("$6+x$", (27,7.5), W);
+label("$6+x$", (27,7.5), S);
-label("$6\sqrt{3}$", (30,7.5),W);
+label("$6\sqrt{3}$", (30,7.5), E);
 label("$T_1$", (10.5,15), N);
 label("$T_2$", (10.5,0), S);
-label("$T_3$", (4.5,11.25),W);
+label("$T_3$", (4.5,11.25), W);
-label("$E$",E, N);
+label("$E$", EE, N);
-label("$F$",F, S);
+label("$F$", F, S);
 </asy>
@@ Line 83: / Line 81: @@
 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 <math>T_1, T_2, T_3</math> be our tangents from the circle to the parallelogram. By the secant power of a point, the power of <math>A = 3 \cdot (3+9) = 36</math>. Then <math>AT_2  = AT_3 = \sqrt{36} = 6</math>. Similarly, the power of <math>C = 16 \cdot (16+9) = 400</math> and  <math>CT_1 = \sqrt{400} = 20</math>. We let <math>BT_3 = BT_1 = x</math> and label the diagram accordingly.
-Notice that because <math>BC = AD, 20+x = 6+DT_2 \implies DT_2 = 14+x</math>. Let <math>O</math> be the center of the circle. Since <math>OT_1</math> and <math>OT_2</math> intersect <math>BC</math> and <math>AD</math>, respectively, at right angles, we have <math>T_2T_2CD</math> 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 <math>D</math> to <math>BC</math> and <math>C</math> to <math>AD</math>, and both are equal to <math>2r</math>. Since <math>T_1E = T_2D</math>, <math>20 - CE = 14+x \implies CE = 6-x</math>. Since <math>CE = DF, DF = 6-x</math> and <math>AF = 6+14+x+6-x = 26</math>. We can now use Pythagorean theorem on <math>\triangle ACF</math>; we have <math>26^2 + (2r)^2 = (3+9+16)^2 \implies 4r^2 = 784-676 \implies 4r^2 = 108 \implies 2r = 6\sqrt{3}</math> and <math>r^2 = 27</math>.
+Notice that because <math>BC = AD, 20+x = 6+DT_2 \implies DT_2 = 14+x</math>. Let <math>O</math> be the center of the circle. Since <math>OT_1</math> and <math>OT_2</math> intersect <math>BC</math> and <math>AD</math>, respectively, at right angles, we have <math>T_2T_1CD</math> 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 <math>D</math> to <math>BC</math> and <math>C</math> to <math>AD</math>, and both are equal to <math>2r</math>. Since <math>T_1E = T_2D</math>, <math>20 - CE = 14+x \implies CE = 6-x</math>. Since <math>CE = DF, DF = 6-x</math> and <math>AF = 6+14+x+6-x = 26</math>. We can now use Pythagorean theorem on <math>\triangle ACF</math>; we have <math>26^2 + (2r)^2 = (3+9+16)^2 \implies 4r^2 = 784-676 \implies 4r^2 = 108 \implies 2r = 6\sqrt{3}</math> and <math>r^2 = 27</math>.
 We know that <math>CD = 6+x</math> because <math>ABCD</math> is a parallelogram. Using Pythagorean theorem on <math>\triangle CDF</math>, <math>(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}</math>. Therefore, base <math>BC = 20 + \frac{9}{2} = \frac{49}{2}</math>. Thus the area of the parallelogram is the base times the height, which is <math>\frac{49}{2} \cdot 6\sqrt{3} = 147\sqrt{3}</math> and the answer is <math>\boxed{150}</math>
@@ Line 94: / Line 92: @@
 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>
-Using POP, it is very clear that <math>PC=20,AQ=AM=6</math>, let <math>BM=BP=x,QD=14+x</math>, using LOC in <math>\triangle{ABP}</math>,<math>x^2+(x+6)^2-2x(x+6)\cos\alpha=36+PQ^2</math>, similarly, use LOC in <math>\triangle{DQC}</math>, getting that <math>(14+x)^2+(6+x)^2-2(6+x)(14+x)\cos\alpha=400+PQ^2</math>. We use the second equation to minus the first equation, getting that <math>28x+196-(2x+12)*14*\cos\alpha=364</math>, we can get <math>\cos\alpha=\frac{2x-12}{2x+12}</math>.
+Using POP, it is very clear that <math>PC=20,AQ=AM=6</math>, let <math>BM=BP=x,QD=14+x</math>, using LOC in <math>\triangle{ABP}</math>,<math>x^2+(x+6)^2-2x(x+6)\cos\alpha=36+PQ^2</math>, similarly, use LOC in <math>\triangle{DQC}</math>, getting that <math>(14+x)^2+(6+x)^2-2(6+x)(14+x)\cos\alpha=400+PQ^2</math>. We use the second equation to minus the first equation, getting that <math>28x+196-(2x+12)\times14\times\cos\alpha=364</math>, we can get <math>\cos\alpha=\frac{2x-12}{2x+12}</math>.
-Now applying LOC in <math>\triangle{ADC}</math>, getting <math>(6+x)^2+400-2(6+x)*20*\frac{2x-12}{2x+12}=(3+9+16)^2</math>, solving this equation to get <math>x=\frac{9}{2}</math>, then <math>\cos\alpha=-\frac{1}{7}</math>, <math>\sin\alpha=\frac{4\sqrt{3}}{7}</math>, the area is <math>\frac{21}{2}*\frac{49}{2}*\frac{4\sqrt{3}}{7}=147\sqrt{3}</math> leads to <math>\boxed{150}</math>
+Now applying LOC in <math>\triangle{ADC}</math>, getting <math>(6+x)^2+(20+x)^2-2(6+x)\times(20+x)\times\frac{2x-12}{2x+12}=(3+9+16)^2</math>, solving this equation to get <math>x=\frac{9}{2}</math>, then <math>\cos\alpha=-\frac{1}{7}</math>, <math>\sin\alpha=\frac{4\sqrt{3}}{7}</math>, the area is <math>\frac{21}{2}\cdot\frac{49}{2}\cdot\frac{4\sqrt{3}}{7}=147\sqrt{3}</math> leads to <math>\boxed{150}</math>
-~bluesoul
+~bluesoul,HarveyZhang
 ==Solution 3==
@@ Line 117: / Line 115: @@
 Hence, <math>\left( 6 + x \right)^2 = 4 r^2 + \left( 6 - x \right)^2</math>.
 This can be simplified as
-\[
+<cmath>\[
 x = r^2 . \hspace{1cm} (1)
-\]
+\]</cmath>
 In <math>\triangle ACB</math>, by applying the law of cosines, we have
-\begin{align*}
+<cmath>\begin{align*}
 AC^2 & = AB^2 + CB^2 - 2 AB \cdot CB \cos B \\
 & = AB^2 + CB^2 + 2 AB \cdot CB \cos A \\
@@ Line 129: / Line 127: @@
 & = \left( 6 + x \right)^2 + \left( 20 + x \right)^2 + 2 \left( 20 + x \right) \left( 6 - x \right) \\
 & = 24 x + 676 .
-\end{align*}
+\end{align*}</cmath>
 Because <math>AC = AP + PQ + QC = 28</math>, we get <math>x = \frac{9}{2}</math>.
@@ Line 135: / Line 133: @@
 Therefore,
-\begin{align*}
+<cmath>\begin{align*}
 {\rm Area} \ ABCD & = CB \cdot EF \\
 & = \left( 20 + x \right) \cdot 2r \\
 & = 147 \sqrt{3} .
-\end{align*}
+\end{align*}</cmath>
 Therefore, the answer is <math>147 + 3 = \boxed{\textbf{(150) }}</math>.
@@ Line 150: / Line 148: @@
 . By the half-base-height formula, <math>[ABC]=r(20+BX)</math>.
-. We can drop altitudes from the center <math>O</math> of <math>\omega</math> to <math>AB</math>, <math>BC</math>, and <math>AC</math>, which have lengths <math>r</math>, <math>r</math>, and <math>\sqrt{r^2-81/4}</math>. Thus, <math>[ABC]=[OAB]+[OBC]+[OAC]=r(BX+13)+14\sqrt{r^2-81/4}</math>.
+. We can drop altitudes from the center <math>O</math> of <math>\omega</math> to <math>AB</math>, <math>BC</math>, and <math>AC</math>, which have lengths <math>r</math>, <math>r</math>, and <math>\sqrt{r^2-\frac{81}{4}}</math>. Thus, <math>[ABC]=[OAB]+[OBC]+[OAC]=r(BX+13)+14\sqrt{r^2-\frac{81}{4}}</math>.
 Equating the two expressions for <math>[ABC]</math> and solving for <math>r</math> yields <math>r=3\sqrt{3}</math>.
@@ Line 159: / Line 157: @@
 ==Solution 5==
-[[File:AIME 2022 I 11.png|700px|right]]
+[[File:AIME-I-2022-11.png|530px|right]]
+Let <math>\omega</math> be the circle, let <math>r</math> be the radius of <math>\omega</math>, and let the points at which <math>\omega</math> is tangent to <math>AB</math>, <math>BC</math>, and <math>AD</math> be <math>H</math>, <math>K</math>, and <math>T</math>, respectively. PoP on <math>A</math> and <math>C</math> with respect to <math>\omega</math> yields <cmath>AT=6, CK=20.</cmath>
+Let <math>TG = AC, CG||AT.</math>
+In  <math>\triangle KGT</math>  <math>KT \perp BC,</math>
+<math>KT = \sqrt{GT^2 – (KC + AT)^2} = 6 \sqrt{3}=2r.</math>
+<math>\angle AOB = 90^{\circ}, OH \perp AB,  OH = r = \frac{KT}{2},</math>
+<cmath>OH^2 = AH \cdot BH \implies  BH = \frac {9}{2}.</cmath>
+Area is <cmath>(BK + KC) \cdot KT = (BH + KC) \cdot 2r = \frac{49}{2} \cdot 6\sqrt{3} = 147 \sqrt{3} \implies 147+3 = \boxed{\textbf{150}}.</cmath>
+'''vladimir.shelomovskii@gmail.com, vvsss'''
+==Solution 6 (Short and Sweet)==
+Let <math>O</math> be the center of the circle. Let points <math>M, N</math> and <math>L</math> be the tangent points of lines <math>BC, AD</math> and <math>AB</math> respectively to the circle. By Power of a Point, <math>({MC})^2=16\cdot{25} \Longrightarrow MC=20</math>. Similarly, <math>({AL})^2=3\cdot{12} \Longrightarrow AL=6</math>. Notice that <math>AL=AN=6</math> since quadrilateral <math>LONA</math> is symmetrical. Let <math>AC</math> intersect <math>MN</math> at <math>I</math>. Then, <math>\bigtriangleup{IMC}</math> is similar to <math>\bigtriangleup{AIN}</math>. Therefore, <math>\frac{CI}{MC}=\frac{AI}{AN}</math>. Let the length of <math>PI=l</math>, then <math>\frac{25-l}{20}=\frac{3+l}{6}</math>. Solving we get <math>l=\frac{45}{13}</math>. Doing the Pythagorean theorem on triangles <math>IMC</math> and <math>AIN</math> for sides <math>MI</math> and <math>IN</math> respectively, we obtain the equation <math>\sqrt{(\frac{280}{13})^2-400} +\sqrt{(\frac{84}{13})^2-36}=MN=2r_1</math> where <math>r_1</math> denotes the radius of the circle. Solving, we get <math>MN=6\sqrt{3}</math>. Additionally, quadrilateral <math>OLBM</math> is symmetrical so <math>OL=OM</math>. Let <math>OL=OM=x</math> and extend a perpendicular foot from <math>B</math> to <math>AD</math> and call it <math>R</math>. Then, <math>\bigtriangleup{ABR}</math> is right with <math>AR=6-x</math>, <math>AB=6+x</math>, and <math>RB=2r_1=MN=6\sqrt{3}</math>. Taking the difference of squares, we get <math>108=24x \Longrightarrow x=\frac{9}{2}</math>. The area of <math>ABCD</math> is <math>MN\cdot{BC}=(20+x)\cdot{MN} \Longrightarrow \frac{49}{2}\cdot{6\sqrt{3}}=147\sqrt{3}</math>. Therefore, the answer is <math>147+3=\boxed{150}</math>
+~[https://artofproblemsolving.com/wiki/index.php/User:Magnetoninja Magnetoninja]
+==Solution 7 (Intuitive, no trig, no weird auxiliary lines)==
+Say that <math>BC</math> is tangent to the circle at <math>X</math> and <math>AD</math> tangent at <math>Y</math>. Also, <math>H</math> is the intersection of <math>XY</math> (diameter) and <math>AC</math> (diagonal). Then by power of a point with given info on <math>A</math> and <math>C</math> we get that <math>AY=6</math> and <math>CX=20</math>. Note that <math>HAY \sim HCX</math>, and since <math>\frac{AY}{CX}=\frac{3}{10}</math> we note that <cmath>\frac{AH}{CH} = \frac{AP+PH}{CQ+QH} = \frac{3+PH}{16+QH} =\frac{AY}{CX}=\frac{3}{10}</cmath>. Since <math>PH+HQ=9</math>, we get that <math>PH=\frac{45}{13}</math> and <math>QH=\frac{72}{13}</math>. This is the length information within the circle.
+The same triangle similarity also means that <math>\frac{YH}{XH}=\frac{3}{10}</math>, so if the radius of the circle is <math>r</math> then we have <math>XH=\frac{20}{13}r</math> and <math>YH = \frac{6}{13}r</math>.
+By power of a point on H, we can figure out <math>r</math>:
+<cmath>XH\cdot YH = PH \cdot QG</cmath>
+<cmath>\frac{20}{13}r \cdot \frac{6}{13}r = \frac{45}{13} \cdot \frac{72}{13}</cmath>
+and we get that <math>r = 3 \sqrt 3</math>. Thus, we have that the height of the parallelogram is <math>2r=6 \sqrt 3</math> and we want to find <math>BC</math>. If <math>AB</math> is tangent to the circle at <math>E</math>, then set <math>a = BX = BE</math>. Using pythagorean theorem, <math>AO^2+BO^2=AB^2</math> and we can plug in diagram values: <cmath>(AY^2+OY^2)+(BX^2+OX)^2=AB^2</cmath> <cmath>(6^2+(3 \sqrt 3)^2) + (a^2+(3 \sqrt 3)^2)=(a+6)^2.</cmath> Solving, we get <math>a=\frac{9}{2}</math>
+Finally, we have <math>[ABCD]=XY \cdot BC = 6 \sqrt 3 \cdot (20+\frac{9}{2}) \rightarrow \boxed{150}</math>
+~ Brocolimanx
+==Solution 8 (Ptolemy's Theorem + Power of Point + Pythagorean Theorem)==
+Let <math>E</math>, <math>F</math>, <math>G</math> be the circle's point of tangency with sides <math>AD</math>, <math>AB</math>, and <math>BC</math>, respectively. Let <math>O</math> be the center of the inscribed circle.
+By Power of a Point, <math>AE^2 = AP \cdot AQ = 3(3+9) = 36</math>, so <math>AE = 6</math>. Similarly, <math>GC^2 = CQ \cdot CP = 16(16+9) = 400</math>, so <math>GC = 20</math>.
+Construct <math>GE</math>, and let <math>I</math> be the point of intersection of <math>GE</math> and <math>AC</math>. <math>GE \perp BC</math> and <math>GE \perp AD</math>. By AA, <math>\triangle IGC \sim \triangle IEA</math>, and we have <math>\frac{AI}{IC} = \frac{AE}{GC} = \frac{3}{10}</math>. We also know <math>AI + IC = AC = 28</math>, so <math>AI = \frac{84}{13}</math> and <math>IC = \frac{280}{13}</math>.
+Using Pythagorean Theorem on <math>\triangle IEA</math> and <math>\triangle CIG</math>, we find that <math>EI = \frac{18\sqrt{3}}{13}</math> and <math>IG = \frac{60\sqrt{3}}{13}</math>. Thus, <math>GE = EI + IG = 6\sqrt{3}</math>, and the radius of the circle is <math>3\sqrt{3}</math>.
+Construct <math>EF</math>, <math>FG</math>. <math>\angle AFO = \angle AEO = 90^{\circ}</math>, so <math>AEOF</math> is cyclic. Similarly, <math>BFOG</math> is cyclic.
+Now, we attempt to set up Ptolemy. Using Pythagorean Theorem on <math>\triangle AEO</math>, we find that <math>AO = 3\sqrt{7}</math>. By Ptolemy's Theorem, <math>(AE)(FO) + (AF)(EO) = (AO)(FE)</math>, from which we have <math>(6)(3\sqrt{3}) + (6)(3\sqrt{3}) = (3\sqrt{7})(FE)</math> and <math>FE = 12\frac{\sqrt{3}}{\sqrt{7}}</math>. From Thales' Circle, <math>\triangle FGE</math> is a right triangle, and <math>EF^2 + FG^2 = GE^2</math>, so <math>FG = \frac{18}{\sqrt{7}}</math>.
+Set <math>BF = BG = s</math>. <math>BO = \sqrt{s^2 + (3\sqrt{3})^2} = \sqrt{s^2+27}</math>, so by Ptolemy's Theorem on <math>BFOG</math>, we have
+<cmath>
+(BF)(GO) + (BG)(FO) = (FG)(BO)
+</cmath>
+<cmath>
+(3\sqrt{3})(s) + (3\sqrt{3})(s) = (\frac{18}{\sqrt{7}})(\sqrt{s^2+27})
+</cmath>
+Solving yields <math>s = \frac{9}{2}</math>.
+We know that <math>BC = BG + GC = 20 + \frac{9}{2} = \frac{49}{2}</math>, so the area of <math>ABCD = (\frac{49}{2})(6\sqrt{3}) = 147\sqrt{3}</math>. The requested answer is <math>147 + 3 = \boxed{150}</math>.
+~ adam_zheng
+==Solution 9==
+Let points <math>E</math>, <math>F</math>, <math>G</math> be the points where lines <math>BC</math>, <math>AB</math>, and <math>AD</math> are tangent to the circle respectively. Then extend line <math>BC</math> to point <math>H</math> such that <math>AH</math> is perpendicular to <math>BC</math>.
+According to Power of a Point, <math>CQ\cdot CP=CE^2\rightarrow 16\cdot 25=CE^2\rightarrow CE=20</math>.
+Similarly, <math>AP\cdot AQ=AF^2\rightarrow 3\cdot 12=AF^2\rightarrow AF=6</math>.
-<math>KC^2 = QC \cdot PC = 20^2, AT^2 = AP \cdot AQ = 6^2.</math>
+(Note that <math>AG=AF=6</math> because they are intersecting tangents from the same circle. Furthermore, <math>HE=AG=6</math> due to a simple upwards translation.)
-Let <math>TG = AC, TG||AT.</math> In right <math>\triangle KGT, KT = \sqrt{GT^2 – (KC + AT)^2} = 6 \sqrt{3}.</math>
-<math>\angle AOB = 90^{\circ}, OH^2 = \frac{KT}{2} = AH \cdot BH, BH = \cdot {9}{2}.</math>
+Therefore, <math>HC=HE+EC=6+20=26</math>, and we already know that <math>AC=28</math>, meaning that we can use the Pythagorean Theorem on <math>\Delta AHC</math> to obtain: <math>AH=6\sqrt{3}</math>.
-Area is <math>(BK + KC) \cdot KT = \frac{49}{2} \cdot 6\sqrt{3} = 147 \sqrt{3}.</math>
+Let <math>HB=k</math>. Then <math>BE=6-HB=6-k</math>. Since they are intersecting tangents from the same circle, <math>BF=BE=6-k</math>.
+Therefore, <math>AB=AF+BF=6+6-k=12-k</math>. We have all the side lengths of <math>\Delta ABH</math>, so applying the Pythagorean Theorem: <math>(6\sqrt{3})^2+k^2=(12-k)^2\rightarrow k=\frac{3}{2}</math>.
+Thus, <math>BC=BE+EC=6-k+20=26-k=\frac{49}{2}</math>.
+<math>[ABCD]=AH\cdot BC=6\sqrt{3}\cdot \frac{49}{2}=147\sqrt{3}</math>, so the answer is <math>\boxed{150}</math>.
+~[https://artofproblemsolving.com/wiki/index.php/User:Sid2012#User sid2012]
 ==Video Solution==
@@ Line 174: / Line 248: @@
 ==Video Solution 2 (Mathematical Dexterity)==
 https://www.youtube.com/watch?v=1nDKQkr9NaU
+== Video Solution 3 by OmegaLearn ==
+https://youtu.be/LpOegT0fKy8?t=740
+~ pi_is_3.14
 ==See Also==

2022 AIME I (Problems • Answer Key • Resources)
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