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

(→‎Solution 1 (No trig))
(→‎Solution 1 (No trig))
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label("$9$", midpoint(P--Q), S);
 
label("$9$", midpoint(P--Q), S);
 
label("$16$", midpoint(Q--C), S);
 
label("$16$", midpoint(Q--C), S);
−
label("$x$", (4.5,11.25), W);
+
label("$x$", (5.5,13.75), W);
 
label("$20$", (20.25,15), N);
 
label("$20$", (20.25,15), N);
 
label("$6$", (5.25,0), S);
 
label("$6$", (5.25,0), S);
Line 68: Line 68:
 
label("$6+x$", (27,7.5), E);
 
label("$6+x$", (27,7.5), E);
 
label("$6\sqrt{3}$", (30,7.5),E);
 
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);
  
 +
</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. By the secant power of a point, the power of <math>A = 3 \cdot (3+9) = 36</math>. Then the length of the tangent from <math>A</math> to the circle is <math>\sqrt{36} = 6</math>. Similarly, the power of <math>D = 16 \cdot (16+9) = 400</math> and the length from <math>D</math> to the tangent of the circle is <math>\sqrt{400} = 20</math>. We let the length from B to the tangents of the circle be <math>x</math> and label the diagram accordingly.
−
 
 
−
 
 
−
</asy>
 
  
 
==Solution 2==
 
==Solution 2==

Revision as of 14:59, 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.

[asy] defaultpen(linewidth(0.6)+fontsize(11)); size(20cm); 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("$3$", 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), E); 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); [/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. By the secant power of a point, the power of $A = 3 \cdot (3+9) = 36$. Then the length of the tangent from $A$ to the circle is $\sqrt{36} = 6$. Similarly, the power of $D = 16 \cdot (16+9) = 400$ and the length from $D$ to the tangent of the circle is $\sqrt{400} = 20$. We let the length from B to the tangents of the circle be $x$ and label the diagram accordingly.

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

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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