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

Revision as of 16:15, 18 May 2023

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$", (18.1,9.7), W); draw(A--B--C--D--cycle); draw(C--A); draw(Circle((10.95,7.45), 7.45)); dot(A^^B^^C^^D^^P^^Q); [/asy]$

Video Solution by Punxsutawney Phil

https://www.youtube.com/watch?v=1m3pqCgwLFE

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)

Denote the center of the circle point $O$ . 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, $16\cdot{25}=({MC})^2$ so $MC$ is $20$ . Similarly, $AL=\sqrt{3\cdot{12}}=6$ . That means $AN$ is also $6$ since quadrilateral $LONA$ is symmetrical. Let $AC$ intersect $MN$ at point $K$ . Then, $\bigtriangleup{KMC}$ is similar to $\bigtriangleup{AKN}$ . Therefore, $\frac{CK}{MC}=\frac{AK}{AN}$ . Let the length of $PK=l$ , then $\frac{25-l}{20}=\frac{3+l}{6}$ . Solving we get $l=\frac{45}{13}$ . Doing the Pythagorean theorem on triangles $KMC$ and $AKN$ for sides $MK$ and $KN$ 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

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

Revision as of 16:14, 18 May 2023 (view source) Magnetoninja (talk \| contribs) (→‎Solution 6 (Short and Sweet)) ← Older edit		Revision as of 16:15, 18 May 2023 (view source) Magnetoninja (talk \| contribs) (→‎Solution 6 (Short and Sweet)) Newer edit →
Line 180:		Line 180:


−	Denote the center of the circle point <math>O</math>. Let points <math>M</math>, <math>N</math>, and <math>L</math> be the tangent points of lines <math>BC</math>, <math>AD</math>, and <math>AB</math> respectively to the circle. By Power of a Point, <math>16\cdot{25}=({MC})^2</math> so <math>MC</math> is <math>20</math>. Similarly, <math>AL~~</math> is~~ \sqrt{3\cdot{12}}=6<math>. That means </math>AN<math> is also </math>6<math> since quadrilateral </math>LONA<math> is symmetrical. Let </math>AC<math> intersect </math>MN<math> at point </math>K<math>. Then, </math>\bigtriangleup{KMC}<math> is similar to </math>\bigtriangleup{AKN}<math>. Therefore, </math>\frac{CK}{MC}=\frac{AK}{AN}<math>. Let the length of </math>PK=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>KMC<math> and </math>AKN<math> for sides </math>MK<math> and </math>KN<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}$	+	Denote the center of the circle point <math>O</math>. Let points <math>M</math>, <math>N</math>, and <math>L</math> be the tangent points of lines <math>BC</math>, <math>AD</math>, and <math>AB</math> respectively to the circle. By Power of a Point, <math>16\cdot{25}=({MC})^2</math> so <math>MC</math> is <math>20</math>. Similarly, <math>AL=\sqrt{3\cdot{12}}=6</math>. That means <math>AN</math> is also <math>6</math> since quadrilateral <math>LONA</math> is symmetrical. Let <math>AC</math> intersect <math>MN</math> at point <math>K</math>. Then, <math>\bigtriangleup{KMC}</math> is similar to <math>\bigtriangleup{AKN}</math>. Therefore, <math>\frac{CK}{MC}=\frac{AK}{AN}</math>. Let the length of <math>PK=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>KMC</math> and <math>AKN</math> for sides <math>MK</math> and <math>KN</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>

	-Magnetoninja		-Magnetoninja

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

Page

Toolbox

Search