AIME 2022 I · 第 11 题

Solution 1 (No trig)

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

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

$$\begin{aligned} AC^2 & = AB^2 + CB^2 - 2 AB \cdot CB \cos B \\ & = AB^2 + CB^2 + 2 AB \cdot CB \cos A \\ & = AB^2 + CB^2 + 2 AB \cdot CB \cdot \frac{AE - BF}{AB} \\ & = AB^2 + CB^2 + 2 CB \left( AE - BF \right) \\ & = \left( 6 + x \right)^2 + \left( 20 + x \right)^2 + 2 \left( 20 + x \right) \left( 6 - x \right) \\ & = 24 x + 676 . \end{aligned}$$ Because $AC = AP + PQ + QC = 28$, we get $x = \frac{9}{2}$. Plugging this into Equation (1), we get $r = 3 \sqrt{3}$.

Therefore,

$$\begin{aligned} {\rm Area} \ ABCD & = CB \cdot EF \\ & = \left( 20 + x \right) \cdot 2r \\ & = 147 \sqrt{3} . \end{aligned}$$ 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

Video Solution 4 by StressedPineapple

https://youtube.com/watch?v=x66z0IMj0as&t=640s