Difference between revisions of "2022 AIME II Problems/Problem 11"
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+ | ==Problem== | ||
+ | Let <math>ABCD</math> be a convex quadrilateral with <math>AB=2, 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== | ||
+ | |||
+ | <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> | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | Denote by <math>M</math> the midpoint of segment <math>BC</math>. | ||
+ | Let points <math>P</math> and <math>Q</math> be on segment <math>AD</math>, such that <math>AP = AB</math> and <math>DQ = DC</math>. | ||
+ | |||
+ | Denote <math>\angle DAM = \alpha</math>, <math>\angle BAD = \beta</math>, <math>\angle BMA = \theta</math>, <math>\angle CMD = \phi</math>. | ||
+ | |||
+ | Denote <math>BM = x</math>. Because <math>M</math> is the midpoint of <math>BC</math>, <math>CM = x</math>. | ||
+ | |||
+ | Because <math>AM</math> is the angle bisector of <math>\angle BAD</math> and <math>AB = AP</math>, <math>\triangle BAM \cong \triangle PAM</math>. | ||
+ | Hence, <math>MP = MB</math> and <math>\angle AMP = \theta</math>. | ||
+ | Hence, <math>\angle MPD = \angle MAP + \angle PMA = \alpha + \theta</math>. | ||
+ | |||
+ | Because <math>DM</math> is the angle bisector of <math>\angle CDA</math> and <math>DC = DQ</math>, <math>\triangle CDM \cong \triangle QDM</math>. | ||
+ | Hence, <math>MQ = MC</math> and <math>\angle DMQ = \phi</math>. | ||
+ | Hence, <math>\angle MQA = \angle MDQ + \angle QMD = \beta + \phi</math>. | ||
+ | |||
+ | Because <math>M</math> is the midpoint of segment <math>BC</math>, <math>MB = MC</math>. | ||
+ | Because <math>MP = MB</math> and <math>MQ = MC</math>, <math>MP = MQ</math>. | ||
+ | |||
+ | Thus, <math>\angle MPD = \angle MQA</math>. | ||
+ | |||
+ | Thus, | ||
+ | <cmath>\[ | ||
+ | \alpha + \theta = \beta + \phi . \hspace{1cm} (1) | ||
+ | \]</cmath> | ||
+ | |||
+ | In <math>\triangle AMD</math>, <math>\angle AMD = 180^\circ - \angle MAD - \angle MDA = 180^\circ - \alpha - \beta</math>. | ||
+ | In addition, <math>\angle AMD = 180^\circ - \angle BMA - \angle CMD = 180^\circ - \theta - \phi</math>. | ||
+ | Thus, | ||
+ | <cmath>\[ | ||
+ | \alpha + \beta = \theta + \phi . \hspace{1cm} (2) | ||
+ | \]</cmath> | ||
+ | |||
+ | Taking <math>(1) + (2)</math>, we get <math>\alpha = \phi</math>. | ||
+ | Taking <math>(1) - (2)</math>, we get <math>\beta = \theta</math>. | ||
+ | |||
+ | Therefore, <math>\triangle ADM \sim \triangle AMB \sim \triangle MDC</math>. | ||
+ | |||
+ | Hence, <math>\frac{AD}{AM} = \frac{AM}{AB}</math> and <math>\frac{AD}{DM} = \frac{DM}{CD}</math>. | ||
+ | Thus, <math>AM = \sqrt{AD \cdot AD} = \sqrt{14}</math> and <math>DM = \sqrt{AD \cdot CD} = \sqrt{21}</math>. | ||
+ | |||
+ | In <math>\triangle ADM</math>, by applying the law of cosines, <math>\cos \angle AMD = \frac{AM^2 + DM^2 - AD^2}{2 AM \cdot DM} = - \frac{1}{\sqrt{6}}</math>. | ||
+ | Hence, <math>\sin \angle AMD = \sqrt{1 - \cos^2 \angle AMD} = \frac{\sqrt{5}}{\sqrt{6}}</math>. | ||
+ | Hence, <math>{\rm Area} \ \triangle ADM = \frac{1}{2} AM \cdot DM \dot \sin \angle AMD = \frac{7 \sqrt{5}}{2}</math>. | ||
+ | |||
+ | Therefore, | ||
+ | <cmath>\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*}</cmath> | ||
+ | |||
+ | Therefore, the square of <math>{\rm Area} \ ABCD</math> is <math>\left( 6 \sqrt{5} \right)^2 = \boxed{\textbf{(180) }}</math>. | ||
+ | |||
+ | ~Steven Chen (www.professorchenedu.com) | ||
+ | |||
+ | ==Solution 3 (Visual)== | ||
+ | [[File:AIME-II-2022-11a.png|300px|right]] | ||
+ | <b><i>Claim</b></i> | ||
+ | |||
+ | In the triangle <math>ABC, AB = 2AC, M</math> is the midpoint of <math>AB. D</math> is the point of intersection of the circumcircle and the bisector of angle <math>A.</math> Then <math>DM = BD.</math> | ||
+ | |||
+ | <b><i> Proof</b></i> | ||
+ | |||
+ | Let <math>A = 2\alpha.</math> Then <math>\angle DBC = \angle DCB = \alpha.</math> | ||
+ | |||
+ | Let <math>E</math> be the intersection point of the perpendicular dropped from <math>D</math> to <math>AB</math> with the circle. | ||
+ | |||
+ | Then the sum of arcs <math>\overset{\Large\frown} {BE} + \overset{\Large\frown}{AC} + \overset{\Large\frown}{CD} = 180^\circ.</math> | ||
+ | <cmath>\overset{\Large\frown} {BE} = 180^\circ – 2\alpha – \overset{\Large\frown}{AC}.</cmath> | ||
+ | |||
+ | Let <math>E'</math> be the point of intersection of the line <math>CM</math> with the circle. | ||
+ | <math>CM</math> is perpendicular to <math>AD, \angle AMC = 90^\circ – \alpha,</math> the sum of arcs <math>\overset{\Large\frown}{A}C + \overset{\Large\frown}{BE'} = 180^\circ – 2\alpha \implies E'</math> coincides with <math>E.</math> | ||
+ | |||
+ | The inscribed angles <math>\angle DEM = \angle DEB, M</math> is symmetric to <math>B</math> with respect to <math>DE, DM = DB.</math> | ||
+ | |||
+ | <b><i> Solution</b></i> | ||
+ | [[File:AIME-II-2022-11b.png|350px|right]] | ||
+ | Let <math>AB' = AB, DC' = DC, B'</math> and <math>C'</math> on <math>AD.</math> | ||
+ | |||
+ | Then <math>AB' = 2, DC' = 3, B'C' = 2 = AB'.</math> | ||
+ | |||
+ | Quadrilateral <math>ABMC'</math> is cyclic. | ||
+ | Let <math>\angle A = 2\alpha.</math> Then <math>\angle MBC' = \angle MC'B = \alpha.</math> | ||
+ | |||
+ | Circle <math>BB'C'C</math> centered at <math>M, BC</math> is its diameter, <math>\angle BC'C = 90^\circ.</math> | ||
+ | <math>\angle DMC' = \angle MC'B,</math> since they both complete <math>\angle MC'C</math> to <math>90^\circ.</math> | ||
+ | |||
+ | <math>\angle MB'A = \angle MC'D,</math> since they are the exterior angles of an isosceles <math>\triangle MB'C'.</math> | ||
+ | <math>\triangle AMB' \sim \triangle MDC'</math> by two angles. | ||
+ | <math>\frac {AB'}{MC'} = \frac {MB'}{DC'}, MC' =\sqrt{AB' \cdot C'D} = \sqrt{6}.</math> | ||
+ | |||
+ | The height dropped from <math>M</math> to <math>AD</math> is <math>\sqrt{MB'^2 - (\frac{B'C'}{2})^2} =\sqrt{6 - 1} = \sqrt{5}.</math> | ||
+ | |||
+ | The areas of triangles <math>\triangle AMB'</math> and <math>\triangle MC'B'</math> are equal to <math>\sqrt{5},</math> area of <math>\triangle MC'D</math> is <math>\frac{3}{2} \sqrt{5}.</math> | ||
+ | |||
+ | <cmath>\triangle AMB' = \triangle AMB, \triangle MC'D = \triangle MCD \implies</cmath> | ||
+ | The area of <math>ABCD</math> is <math>(1 + 2 + 3) \sqrt{5} = 6\sqrt{5} \implies 6^2 \cdot 5 = \boxed{180}.</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Video Solution by The Power of Logic== | ||
+ | https://youtu.be/giLyWHKFr1I | ||
+ | |||
+ | ==See Also== | ||
+ | {{AIME box|year=2022|n=II|num-b=10|num-a=12}} | ||
+ | {{MAA Notice}} |
Revision as of 08:18, 30 March 2023
Contents
Problem
Let be a convex quadrilateral with and such that the bisectors of acute angles and intersect at the midpoint of Find the square of the area of
Solution 1
According to the problem, we have , , , , and
Because is the midpoint of , we have , so:
Then, we can see that is an isosceles triangle with
Therefore, we could start our angle chasing: .
This is when we found that points , , , and are on a circle. Thus, . This is the time we found that .
Thus,
Point is the midpoint of , and . .
The area of this quadrilateral is the sum of areas of triangles:
Finally, the square of the area is
Solution 2
Denote by the midpoint of segment . Let points and be on segment , such that and .
Denote , , , .
Denote . Because is the midpoint of , .
Because is the angle bisector of and , . Hence, and . Hence, .
Because is the angle bisector of and , . Hence, and . Hence, .
Because is the midpoint of segment , . Because and , .
Thus, .
Thus,
In , . In addition, . Thus,
Taking , we get . Taking , we get .
Therefore, .
Hence, and . Thus, and .
In , by applying the law of cosines, . Hence, . Hence, .
Therefore,
Therefore, the square of is .
~Steven Chen (www.professorchenedu.com)
Solution 3 (Visual)
Claim
In the triangle is the midpoint of is the point of intersection of the circumcircle and the bisector of angle Then
Proof
Let Then
Let be the intersection point of the perpendicular dropped from to with the circle.
Then the sum of arcs
Let be the point of intersection of the line with the circle. is perpendicular to the sum of arcs coincides with
The inscribed angles is symmetric to with respect to
Solution
Let and on
Then
Quadrilateral is cyclic. Let Then
Circle centered at is its diameter, since they both complete to
since they are the exterior angles of an isosceles by two angles.
The height dropped from to is
The areas of triangles and are equal to area of is
The area of is
vladimir.shelomovskii@gmail.com, vvsss
Video Solution by The Power of Logic
See Also
2022 AIME II (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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.