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

Revision as of 11:58, 7 February 2023

Problem

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 $ABCD$ .

Vishal is gay

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)

Claim

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 $DM = BD.$

Proof

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 $DE, DM = DB.$

Solution

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

Video Solution by The Power of Logic

https://youtu.be/giLyWHKFr1I

@@ Line 2: / Line 2: @@
 Let <math>ABCD</math> be a convex quadrilateral with <math>AB=2</math>, <math>AD=7</math>, and <math>CD=3</math> such that the bisectors of acute angles <math>\angle{DAB}</math> and <math>\angle{ADC}</math> intersect at the midpoint of <math>\overline{BC}</math>. Find the square of the area of <math>ABCD</math>.
-==Solution 1==
+Vishal is gay
-<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 <math>AB=AB'=2</math>, <math>DC=DC'=3</math>, <math>MB=MB'</math>, <math>MC=MC'</math>, and <math>B'C'=7-2-3=2</math>
-Because <math>M</math> is the midpoint of <math>BC</math>, we have <math>BM=MC</math>, so: <cmath>MB=MB'=MC'=MC.</cmath>
-Then, we can see that <math>\bigtriangleup{MB'C'}</math> is an isosceles triangle with <math>MB'=MC'</math>
-Therefore, we could start our angle chasing: <math>\angle{MB'C'}=\angle{MC'B'}=180^\circ-\angle{MC'D}=180^\circ-\angle{MCD}</math>.
-This is when we found that points <math>M</math>, <math>C</math>, <math>D</math>, and <math>B'</math> are on a circle. Thus, <math>\angle{BMB'}=\angle{CDC'} \Rightarrow \angle{B'MA}=\angle{C'DM}</math>. This is the time we found that <math>\bigtriangleup{AB'M} \sim \bigtriangleup{MC'D}</math>.
-Thus, <math>\frac{AB'}{B'M}=\frac{MC'}{C'D} \Longrightarrow (B'M)^2=AB' \cdot C'D = 6</math>
-Point <math>H</math> is the midpoint of <math>B'C'</math>, and <math>MH \perp AD</math>. <math>B'H=HC'=1 \Longrightarrow MH=\sqrt{B'M^2-B'H^2}=\sqrt{6-1}=\sqrt{5}</math>.
-The area of this quadrilateral is the sum of areas of triangles: <cmath>S_{\bigtriangleup{ABM}}+S_{\bigtriangleup{AB'M}}+S_{\bigtriangleup{CDM}}+S_{\bigtriangleup{CD'M}}+S_{\bigtriangleup{B'C'M}}</cmath>
-<cmath>=S_{\bigtriangleup{AB'M}}\cdot 2 + S_{\bigtriangleup{B'C'M}} + S_{\bigtriangleup{C'DM}}\cdot 2</cmath>
-<cmath>=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</cmath>
-<cmath>=2\sqrt{5}+\sqrt{5}+3\sqrt{5}=6\sqrt{5}</cmath>
-Finally, the square of the area is <math>(6\sqrt{5})^2=\boxed{180}</math>
-~DSAERF-CALMIT (https://binaryphi.site)
 ==Solution 2==

2022 AIME II (Problems • Answer Key • Resources)
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