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''' | ||
+ | |||
+ | ==Solution 4 (THINK OUTSIDE THE BOX)== | ||
+ | |||
+ | Extend <math>AB</math> and <math>CD</math> so they intersect at a point <math>X</math>. Then note that <math>M</math> is the incenter of <math>\triangle{XAD}</math>, implying that <math>M</math> is on the angle bisector of <math>X</math>. Now because <math>XM</math> is both an angle bisector and a median of <math>\triangle{XBC}</math>, <math>\triangle{XBC}</math> is isosceles. Then we can start angle chasing: | ||
+ | |||
+ | Let <math>\angle{BAM}=a, \angle{CDM}=b,</math> and <math>\angle{XBC}=c</math>. Then <math>\angle{AMD}=\pi-(a+b), \angle{ABM}=\pi-c, \angle{DCM}=\pi-c</math>, implying that <math>\angle{BMA}+\angle{CMD}=a+b</math>, implying that <math>2c-(a+b)=(a+b)</math>, or that <math>c=a+b</math>. Substituting this into the rest of the diagram, we find that <math>\triangle{BMA} \sim \triangle{CDM} \sim \triangle{MDA}</math>. | ||
+ | |||
+ | Then <math>\frac{AB}{BM}=\frac{MC}{CD}</math>, or <math>BM=CM=\sqrt{6}</math>. Moreover, <math>\frac{AB}{AM}=\frac{AM}{AD}</math>, or <math>AM=\sqrt{14}</math>. Similarly, <math>\frac{CD}{MD}=\frac{MD}{AD}</math>, or <math>DM=\sqrt{21}</math>. Then using Law of Cosines on <math>\triangle{AMD}</math>, to get that <math>cos\angle{AMD}=-\frac{\sqrt{6}}{6}</math>, or <math>sin\angle{AMD}=\frac{\sqrt{30}}{6}</math>. | ||
+ | |||
+ | We finish by using the formula <math>K=\frac{1}{2}absinC</math>, as follows: | ||
+ | |||
+ | <math>[ABCD]=[ABM]+[CDM]+[ADM]=\frac{\frac{\sqrt{30}}{6}(2\sqrt{6}+3\sqrt{6}+7\sqrt{6})}{2}=6\sqrt{5}</math>. | ||
+ | |||
+ | <math>(6\sqrt{5})^2=\boxed{180}</math>. | ||
+ | |||
+ | -dragoon | ||
+ | |||
+ | ==Solution 5 (incircle)== | ||
+ | [[File:2022_AIME_II_Problem_11_Diagram.png|300px|right]] | ||
+ | |||
+ | As shown in paragraph one of solution 4, extending <math>AB</math> and <math>CD</math> to <math>X</math>, we realize that <math>\triangle{XBC}</math> is isosceles, thus <math>XM \perp BC</math>. Let <math>XB = XC = x</math>. And, midpoint <math>M</math> is the incenter of <math>\triangle{XAD}</math>. Construct perpendiculars <math>ME, MF, MG</math> to sides <math>AD, AX, DX</math> respectively (constructing the radii of the incircle). Let <math>EM = FM = GM = r</math>. The semiperimeter <math>s = \frac{2x + 2 + 3 + 7}{2} = x+6</math>. Since <math>FX</math> is the tangent off the incircle, <math>FX = s - AD = x-1</math>. So, <math>BF = BX - FX = 1</math>. Because <math>\triangle{BFM} \sim \triangle{MFX}</math>, | ||
+ | |||
+ | <cmath>\frac{BF}{FM} = \frac{FM}{FX} \implies {FM}^2 = BF \cdot FX = x - 1 \implies r^2 = x - 1</cmath>. | ||
+ | |||
+ | By Heron's formula and the inradius area formula, | ||
+ | |||
+ | <cmath>(x+6)r = \sqrt{(x+6)\cdot 4 \cdot 3 \cdot (x-1)} \implies (x+6)r^2 = 12(x-1) \implies (x+6)(x-1) = 12(x-1) \implies x=6</cmath> | ||
+ | |||
+ | Then, <math>r^2 = x - 1 = 5 \implies r = \sqrt{5}</math>. Finally, | ||
+ | |||
+ | <cmath>[ABCD] = [ABM] + [CDM] + [AMD] = \frac{AB \cdot FM}{2} + \frac{CD \cdot GM}{2} + \frac{AD \cdot EM}{2} = \frac{2r}{2} + \frac{3r}{2} + \frac{7r}{2} = 6r = 6\sqrt{5}</cmath> | ||
+ | |||
+ | Thus, our answer is <math>(6\sqrt{5})^{2} = \boxed{180}</math> | ||
+ | |||
+ | ===Remark=== | ||
+ | <math>B</math> and <math>C</math> are the tangent points of the X-mixtilinear incircle (of <math>\triangle{XAD}</math>). This may be useful, but I haven't looked into it. | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Crazyvideogamez CrazyVideoGamez] | ||
+ | |||
+ | ==Solution 6 (bash)== | ||
+ | Let the midpoint of <math>BC</math> be <math>M</math>. Angle-chase and observe that <math>\Delta AMD~\Delta ABM~\Delta MCD</math>. Let <math>BM=CM=a</math> and <math>AM=x</math> and <math>DM=y</math>. As a result of this similarity, we write | ||
+ | |||
+ | <cmath>\dfrac2a=\dfrac a3,</cmath> | ||
+ | |||
+ | which gives <math>a=\sqrt 6</math>. Similarly, we write | ||
+ | |||
+ | <cmath>\dfrac2x=\dfrac x7</cmath> | ||
+ | |||
+ | and | ||
+ | |||
+ | <cmath>\dfrac3y=\dfrac y7</cmath> | ||
+ | |||
+ | to get <math>x=\sqrt{14}</math> and <math>y=\sqrt{21}</math>. | ||
+ | |||
+ | We now have all required side lengths; we can find the area of <math>\Delta AMD</math> with Heron's formula. Doing so yields <math>\dfrac72\sqrt5</math>. We could also bash out the areas of the other two triangles since we know all their side lengths (this is what I did :sob:), but a more intelligent method is to recall the triangles' similarity. The ratio of similarlity between <math>\Delta AMD</math> and <math>\Delta ABM</math> is <math>\dfrac{\sqrt{14}}7=\sqrt{\dfrac27}</math>, and between <math>\Delta AMD</math> and <math>\Delta MCD</math> is <math>\dfrac{\sqrt{21}}7=\sqrt{\dfrac37}</math>. Thus, the area ratios are <math>\dfrac27</math> and <math>\dfrac37</math>, respectively, so adding together we have <math>\dfrac27+\dfrac37=\dfrac57</math>. Multiplying this by our <math>\dfrac72\sqrt5</math>, we have <math>\dfrac52\sqrt5</math> as their total area. Adding this to our original area, we have <math>\dfrac52\sqrt5+\dfrac72\sqrt5=\sqrt5\left(\dfrac52+\dfrac72\right)=\sqrt5\left(\dfrac{12}2\right)=6\sqrt5</math>. | ||
+ | |||
+ | The square of this is <math>\boxed{180}</math>. | ||
+ | |||
+ | ~~Technodoggo | ||
+ | |||
+ | ==Solution 7 (similar to 4)== | ||
+ | As in solution 4, let <math>X=AB\cap CD</math>, so <math>M</math> is the incenter of <math>ADX</math> and <math>XB=XC</math>. Let <math>XB=XC=x</math>. Then the normalized barycentric coordinates of <math>B</math>, <math>C</math>, and <math>M</math> with respect to <math>ADX</math> are <math>\left[\frac{x}{x+2}:0:\frac{2}{x+2}\right]</math>, <math>\left[0:\frac{x}{x+3}:\frac{3}{x+3}\right]</math>, and <math>\left[\frac{x+3}{2x+12}:\frac{x+2}{2x+12}:\frac{7}{2x+12}\right]</math>. So we have <math>\frac{1}{2}\frac{x}{x+2}=\frac{x+3}{2x+12}</math> giving <math>x=6</math>. The sidelengths of <math>ADX</math> are thus <math>AD=7</math>, <math>AX=8</math>, and <math>DX=9</math> giving <math>[ADX]=12\sqrt 5</math>. Also, we have <math>[BCX]=\frac{6}{8}\cdot\frac{6}{9}[ADX]=6\sqrt 5</math> so that <math>[ABCD]=[ADX]-[BCX]=6\sqrt 5</math>. The area squared is thus <math>\boxed{180}</math>. | ||
+ | ~~ golue3120 | ||
+ | |||
+ | ==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}} |
Latest revision as of 18:26, 8 February 2024
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
Solution 4 (THINK OUTSIDE THE BOX)
Extend and
so they intersect at a point
. Then note that
is the incenter of
, implying that
is on the angle bisector of
. Now because
is both an angle bisector and a median of
,
is isosceles. Then we can start angle chasing:
Let and
. Then
, implying that
, implying that
, or that
. Substituting this into the rest of the diagram, we find that
.
Then , or
. Moreover,
, or
. Similarly,
, or
. Then using Law of Cosines on
, to get that
, or
.
We finish by using the formula , as follows:
.
.
-dragoon
Solution 5 (incircle)
As shown in paragraph one of solution 4, extending and
to
, we realize that
is isosceles, thus
. Let
. And, midpoint
is the incenter of
. Construct perpendiculars
to sides
respectively (constructing the radii of the incircle). Let
. The semiperimeter
. Since
is the tangent off the incircle,
. So,
. Because
,
.
By Heron's formula and the inradius area formula,
Then, . Finally,
Thus, our answer is
Remark
and
are the tangent points of the X-mixtilinear incircle (of
). This may be useful, but I haven't looked into it.
Solution 6 (bash)
Let the midpoint of be
. Angle-chase and observe that
. Let
and
and
. As a result of this similarity, we write
which gives . Similarly, we write
and
to get and
.
We now have all required side lengths; we can find the area of with Heron's formula. Doing so yields
. We could also bash out the areas of the other two triangles since we know all their side lengths (this is what I did :sob:), but a more intelligent method is to recall the triangles' similarity. The ratio of similarlity between
and
is
, and between
and
is
. Thus, the area ratios are
and
, respectively, so adding together we have
. Multiplying this by our
, we have
as their total area. Adding this to our original area, we have
.
The square of this is .
~~Technodoggo
Solution 7 (similar to 4)
As in solution 4, let , so
is the incenter of
and
. Let
. Then the normalized barycentric coordinates of
,
, and
with respect to
are
,
, and
. So we have
giving
. The sidelengths of
are thus
,
, and
giving
. Also, we have
so that
. The area squared is thus
.
~~ golue3120
Video Solution by The Power of Logic
See Also
2022 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.