2022 AIME II Problems/Problem 11

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Let $ABCD$ be a convex quadrilateral with $AB=2, AD=7,$ and $CD=3$ such that the bisectors of acute angles $\angle{DAB}$ and $\angle{ADC}$ intersect at the midpoint of $\overline{BC}.$ Find the square of the area of $ABCD.$

Solution 1

[asy]  defaultpen(fontsize(12)+0.6); size(300);  pair A,B,C,D,M,H; real xb=71, xd=121; A=origin; D=(7,0); B=2*dir(xb); C=3*dir(xd)+D; M=(B+C)/2; H=foot(M,A,D); path c=CR(D,3); pair A1=bisectorpoint(D,A,B), D1=bisectorpoint(C,D,A), Bp=IP(CR(A,2),A--H), Cp=IP(CR(D,3),D--H);  draw(B--A--D--C--B); draw(A--M--D^^M--H^^Bp--M--Cp, gray+0.4); draw(rightanglemark(A,H,M,5));  dot("$A$",A,SW); dot("$D$",D,SE); dot("$B$",B,NW); dot("$C$",C,NE); dot("$M$",M,up); dot("$H$",H,down); dot("$B'$",Bp,down); dot("$C'$",Cp,down); [/asy]

According to the problem, we have $AB=AB'=2$, $DC=DC'=3$, $MB=MB'$, $MC=MC'$, and $B'C'=7-2-3=2$

Because $M$ is the midpoint of $BC$, we have $BM=MC$, so: \[MB=MB'=MC'=MC.\]

Then, we can see that $\bigtriangleup{MB'C'}$ is an isosceles triangle with $MB'=MC'$

Therefore, we could start our angle chasing: $\angle{MB'C'}=\angle{MC'B'}=180^\circ-\angle{MC'D}=180^\circ-\angle{MCD}$.

This is when we found that points $M$, $C$, $D$, and $B'$ are on a circle. Thus, $\angle{BMB'}=\angle{CDC'} \Rightarrow \angle{B'MA}=\angle{C'DM}$. This is the time we found that $\bigtriangleup{AB'M} \sim \bigtriangleup{MC'D}$.

Thus, $\frac{AB'}{B'M}=\frac{MC'}{C'D} \Longrightarrow (B'M)^2=AB' \cdot C'D = 6$

Point $H$ is the midpoint of $B'C'$, and $MH \perp AD$. $B'H=HC'=1 \Longrightarrow MH=\sqrt{B'M^2-B'H^2}=\sqrt{6-1}=\sqrt{5}$.

The area of this quadrilateral is the sum of areas of triangles: \[S_{\bigtriangleup{ABM}}+S_{\bigtriangleup{AB'M}}+S_{\bigtriangleup{CDM}}+S_{\bigtriangleup{CD'M}}+S_{\bigtriangleup{B'C'M}}\] \[=S_{\bigtriangleup{AB'M}}\cdot 2 + S_{\bigtriangleup{B'C'M}} + S_{\bigtriangleup{C'DM}}\cdot 2\] \[=2 \cdot \frac{1}{2} \cdot AB' \cdot MH + \frac{1}{2} \cdot B'C' \cdot MH + 2 \cdot \frac{1}{2} \cdot C'D \cdot MH\] \[=2\sqrt{5}+\sqrt{5}+3\sqrt{5}=6\sqrt{5}\]

Finally, the square of the area is $(6\sqrt{5})^2=\boxed{180}$

Solution 2

Denote by $M$ the midpoint of segment $BC$. Let points $P$ and $Q$ be on segment $AD$, such that $AP = AB$ and $DQ = DC$.

Denote $\angle DAM = \alpha$, $\angle BAD = \beta$, $\angle BMA = \theta$, $\angle CMD = \phi$.

Denote $BM = x$. Because $M$ is the midpoint of $BC$, $CM = x$.

Because $AM$ is the angle bisector of $\angle BAD$ and $AB = AP$, $\triangle BAM \cong \triangle PAM$. Hence, $MP = MB$ and $\angle AMP = \theta$. Hence, $\angle MPD = \angle MAP + \angle PMA = \alpha + \theta$.

Because $DM$ is the angle bisector of $\angle CDA$ and $DC = DQ$, $\triangle CDM \cong \triangle QDM$. Hence, $MQ = MC$ and $\angle DMQ = \phi$. Hence, $\angle MQA = \angle MDQ + \angle QMD = \beta + \phi$.

Because $M$ is the midpoint of segment $BC$, $MB = MC$. Because $MP = MB$ and $MQ = MC$, $MP = MQ$.

Thus, $\angle MPD = \angle MQA$.

Thus, \[ \alpha + \theta = \beta + \phi . \hspace{1cm} (1) \]

In $\triangle AMD$, $\angle AMD = 180^\circ - \angle MAD - \angle MDA = 180^\circ - \alpha - \beta$. In addition, $\angle AMD = 180^\circ - \angle BMA - \angle CMD = 180^\circ - \theta - \phi$. Thus, \[ \alpha + \beta = \theta + \phi . \hspace{1cm} (2) \]

Taking $(1) + (2)$, we get $\alpha = \phi$. Taking $(1) - (2)$, we get $\beta = \theta$.

Therefore, $\triangle ADM \sim \triangle AMB \sim \triangle MDC$.

Hence, $\frac{AD}{AM} = \frac{AM}{AB}$ and $\frac{AD}{DM} = \frac{DM}{CD}$. Thus, $AM = \sqrt{AD \cdot AD} = \sqrt{14}$ and $DM = \sqrt{AD \cdot CD} = \sqrt{21}$.

In $\triangle ADM$, by applying the law of cosines, $\cos \angle AMD  = \frac{AM^2 + DM^2 - AD^2}{2 AM \cdot DM} = - \frac{1}{\sqrt{6}}$. Hence, $\sin \angle AMD = \sqrt{1 - \cos^2 \angle AMD} = \frac{\sqrt{5}}{\sqrt{6}}$. Hence, ${\rm Area} \ \triangle ADM = \frac{1}{2} AM \cdot DM \dot \sin \angle AMD = \frac{7 \sqrt{5}}{2}$.

Therefore, \begin{align*} {\rm Area} \ ABCD & = {\rm Area} \ \triangle AMD + {\rm Area} \ \triangle ABM + {\rm Area} \ \triangle MCD \\ & = {\rm Area} \ \triangle AMD \left( 1 + \left( \frac{AM}{AD} \right)^2 + \left( \frac{MD}{AD} \right)^2 \right) \\ & = 6 \sqrt{5} . \end{align*}

Therefore, the square of ${\rm Area} \ ABCD$ is $\left( 6 \sqrt{5} \right)^2 = \boxed{\textbf{(180) }}$.

~Steven Chen (www.professorchenedu.com)

Solution 3 (Visual)



In the triangle $ABC, AB = 2AC, M$ is the midpoint of $AB. D$ is the point of intersection of the circumcircle and the bisector of angle $A.$ Then $DM = BD.$


Let $A = 2\alpha.$ Then $\angle DBC = \angle DCB = \alpha.$

Let $E$ be the intersection point of the perpendicular dropped from $D$ to $AB$ with the circle.

Then the sum of arcs $\overset{\Large\frown} {BE} + \overset{\Large\frown}{AC} + \overset{\Large\frown}{CD} = 180^\circ.$ \[\overset{\Large\frown} {BE} = 180^\circ – 2\alpha – \overset{\Large\frown}{AC}.\]

Let $E'$ be the point of intersection of the line $CM$ with the circle. $CM$ is perpendicular to $AD, \angle AMC = 90^\circ – \alpha,$ the sum of arcs $\overset{\Large\frown}{A}C +  \overset{\Large\frown}{BE'} = 180^\circ – 2\alpha \implies E'$ coincides with $E.$

The inscribed angles $\angle DEM = \angle DEB, M$ is symmetric to $B$ with respect to $DE, DM = DB.$



Let $AB' = AB, DC' = DC, B'$ and $C'$ on $AD.$

Then $AB' = 2, DC' = 3, B'C' = 2 = AB'.$

Quadrilateral $ABMC'$ is cyclic. Let $\angle A = 2\alpha.$ Then $\angle MBC' = \angle MC'B = \alpha.$

Circle $BB'C'C$ centered at $M, BC$ is its diameter, $\angle BC'C = 90^\circ.$ $\angle DMC' = \angle MC'B,$ since they both complete $\angle MC'C$ to $90^\circ.$

$\angle MB'A = \angle MC'D,$ since they are the exterior angles of an isosceles $\triangle MB'C'.$ $\triangle AMB' \sim \triangle MDC'$ by two angles. $\frac {AB'}{MC'} = \frac {MB'}{DC'}, MC' =\sqrt{AB' \cdot C'D} = \sqrt{6}.$

The height dropped from $M$ to $AD$ is $\sqrt{MB'^2 - (\frac{B'C'}{2})^2} =\sqrt{6 - 1} = \sqrt{5}.$

The areas of triangles $\triangle AMB'$ and $\triangle MC'B'$ are equal to $\sqrt{5},$ area of $\triangle MC'D$ is $\frac{3}{2} \sqrt{5}.$

\[\triangle AMB' = \triangle AMB, \triangle MC'D = \triangle MCD \implies\] The area of $ABCD$ is $(1 + 2 + 3) \sqrt{5} = 6\sqrt{5} \implies 6^2 \cdot 5 = \boxed{180}.$

vladimir.shelomovskii@gmail.com, vvsss


Extend $AB$ and $CD$ so they intersect at a point $X$. Then note that $M$ is the incenter of $\triangle{XAD}$, implying that $M$ is on the angle bisector of $X$. Now because $XM$ is both an angle bisector and a median of $\triangle{XBC}$, $\triangle{XBC}$ is isosceles. Then we can start angle chasing:

Let $\angle{BAM}=a, \angle{CDM}=b,$ and $\angle{XBC}=c$. Then $\angle{AMD}=\pi-(a+b), \angle{ABM}=\pi-c, \angle{DCM}=\pi-c$, implying that $\angle{BMA}+\angle{CMD}=a+b$, implying that $2c-(a+b)=(a+b)$, or that $c=a+b$. Substituting this into the rest of the diagram, we find that $\triangle{BMA} \sim \triangle{CDM} \sim \triangle{MDA}$.

Then $\frac{AB}{BM}=\frac{MC}{CD}$, or $BM=CM=\sqrt{6}$. Moreover, $\frac{AB}{AM}=\frac{AM}{AD}$, or $AM=\sqrt{14}$. Similarly, $\frac{CD}{MD}=\frac{MD}{AD}$, or $DM=\sqrt{21}$. Then using Law of Cosines on $\triangle{AMD}$, to get that $cos\angle{AMD}=-\frac{\sqrt{6}}{6}$, or $sin\angle{AMD}=\frac{\sqrt{30}}{6}$.

We finish by using the formula $K=\frac{1}{2}absinC$, as follows:




Video Solution by The Power of Logic


See Also

2022 AIME II (ProblemsAnswer KeyResources)
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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