Difference between revisions of "2021 AIME II Problems/Problem 12"

Latest revision as of 16:33, 12 January 2024

Problem

A convex quadrilateral has area $30$ and side lengths $5, 6, 9,$ and $7,$ in that order. Denote by $\theta$ the measure of the acute angle formed by the diagonals of the quadrilateral. Then $\tan \theta$ can be written in the form $\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ .

Solution 1 (Sines and Cosines)

Since we are asked to find $\tan \theta$ , we can find $\sin \theta$ and $\cos \theta$ separately and use their values to get $\tan \theta$ . We can start by drawing a diagram. Let the vertices of the quadrilateral be $A$ , $B$ , $C$ , and $D$ . Let $AB = 5$ , $BC = 6$ , $CD = 9$ , and $DA = 7$ . Let $AX = a$ , $BX = b$ , $CX = c$ , and $DX = d$ . We know that $\theta$ is the acute angle formed between the intersection of the diagonals $AC$ and $BD$ . $[asy] unitsize(4cm); pair A,B,C,D,X; A = (0,0); B = (1,0); C = (1.25,-1); D = (-0.75,-0.75); draw(A--B--C--D--cycle,black+1bp); X = intersectionpoint(A--C,B--D); draw(A--C); draw(B--D); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); dot(X); label("$X$",X,S); label("$5$",(A+B)/2,N); label("$6$",(B+C)/2,E); label("$9$",(C+D)/2,S); label("$7$",(D+A)/2,W); label("$\theta$",X,2.5E); label("$a$",(A+X)/2,NE); label("$b$",(B+X)/2,NW); label("$c$",(C+X)/2,SW); label("$d$",(D+X)/2,SE); [/asy]$ We are given that the area of quadrilateral $ABCD$ is $30$ . We can express this area using the areas of triangles $AXB$ , $BXC$ , $CXD$ , and $DXA$ . Since we want to find $\sin \theta$ and $\cos \theta$ , we can represent these areas using $\sin \theta$ as follows: $\begin{align*} 30 &=[ABCD] \\ &=[AXB] + [BXC] + [CXD] + [DXA] \\ &=\frac{1}{2} ab \sin (\angle AXB) + \frac{1}{2} bc \sin (\angle BXC) + \frac{1}{2} cd \sin (\angle CXD) + \frac{1}{2} da \sin (\angle AXD) \\ &=\frac{1}{2} ab \sin (180^\circ - \theta) + \frac{1}{2} bc \sin (\theta) + \frac{1}{2} cd \sin (180^\circ - \theta) + \frac{1}{2} da \sin (\theta). \end{align*}$ We know that $\sin (180^\circ - \theta) = \sin \theta$ . Therefore it follows that: $\begin{align*} 30 &=\frac{1}{2} ab \sin (180^\circ - \theta) + \frac{1}{2} bc \sin (\theta) + \frac{1}{2} cd \sin (180^\circ - \theta) + \frac{1}{2} da \sin (\theta) \\ &=\frac{1}{2} ab \sin (\theta) + \frac{1}{2} bc \sin (\theta) + \frac{1}{2} cd \sin (\theta) + \frac{1}{2} da \sin (\theta) \\ &=\frac{1}{2}\sin\theta (ab + bc + cd + da). \end{align*}$ From here we see that $\sin \theta = \frac{60}{ab + bc + cd + da}$ . Now we need to find $\cos \theta$ . Using the Law of Cosines on each of the four smaller triangles, we get following equations: $\begin{align*} 5^2 &= a^2 + b^2 - 2ab\cos(180^\circ-\theta), \\ 6^2 &= b^2 + c^2 - 2bc\cos \theta, \\ 9^2 &= c^2 + d^2 - 2cd\cos(180^\circ-\theta), \\ 7^2 &= d^2 + a^2 - 2da\cos \theta. \end{align*}$ We know that $\cos (180^\circ - \theta) = -\cos \theta$ for all $\theta$ . We can substitute this value into our equations to get: $\begin{align*} 5^2 &= a^2 + b^2 + 2ab\cos \theta, &&(1) \\ 6^2 &= b^2 + c^2 - 2bc\cos \theta, &&(2) \\ 9^2 &= c^2 + d^2 + 2cd\cos \theta, &&(3) \\ 7^2 &= d^2 + a^2 - 2da\cos \theta. &&(4) \end{align*}$ If we subtract $(2)+(4)$ from $(1)+(3)$ , the squared terms cancel, leaving us with: $\begin{align*} 5^2 + 9^2 - 6^2 - 7^2 &= 2ab \cos \theta + 2bc \cos \theta + 2cd \cos \theta + 2da \cos \theta \\ 21 &= 2\cos \theta (ab + bc + cd + da). \end{align*}$ From here we see that $\cos \theta = \frac{21/2}{ab + bc + cd + da}$ .

Since we have figured out $\sin \theta$ and $\cos \theta$ , we can calculate $\tan \theta$ : $\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{60}{ab + bc + cd + da}}{\frac{21/2}{ab + bc + cd + da}} = \frac{60}{21/2} = \frac{120}{21} = \frac{40}{7}.$ Therefore our answer is $40 + 7 = \boxed{047}$ .

~ Steven Chen (www.professorchenedu.com)

~ my_aops_lessons

Solution 2 (Right Triangles)

In convex quadrilateral $ABCD,$ let $AB=5,BC=6,CD=9,$ and $DA=7.$ Let $A'$ and $C'$ be the feet of the perpendiculars from $A$ and $C,$ respectively, to $\overline{BD}.$ We obtain the following diagram: $[asy] /* Made by MRENTHUSIASM */ size(500); pair A, B, C, D, P, A1, C1; B = origin; D = (3*sqrt(32498*(29400*sqrt(47)+312523))/32498,0); A = intersectionpoints(Circle(B,5),Circle(D,7))[0]; C = intersectionpoints(Circle(B,6),Circle(D,9))[1]; P = intersectionpoint(A--C,B--D); A1 = foot(A,B,D); C1 = foot(C,B,D); markscalefactor=3/160; draw(rightanglemark(A,A1,D),red); draw(rightanglemark(C,C1,B),red); dot("$A$",A,1.5*dir(aCos(7/sqrt(1649)))); dot("$B$",B,1.5*W); dot("$C$",C,1.5*dir(180+aCos(7/sqrt(1649)))); dot("$D$",D,1.5*E); dot("$E$",P,dir(180-(180-aCos(7/sqrt(1649)))/2)); dot("$A'$",A1,dir(-75)); dot("$C'$",C1,N); label("$\theta$",P,dir(180+aCos(7/sqrt(1649))/2),red); draw(A--A1^^C--C1,dashed); draw(A--B--C--D--cycle^^A--C^^B--D); [/asy]$ Let $BC'=p,C'E=q,EA'=r,A'D=s,AA'=h_1,$ and $CC'=h_2.$ We apply the Pythagorean Theorem to right triangles $\triangle ABA',\triangle BCC',\triangle CDC',$ and $\triangle DAA',$ respectively: $\begin{array}{ccccccccccccccccc} (p+q+r)^2&+&h_1^2&=&5^2, &&&&&&&&&&&&\hspace{36mm}(1) \\ [1ex] p^2&+&h_2^2&=&6^2, &&&&&&&&&&&&\hspace{36mm}(2) \\ [1ex] (q+r+s)^2&+&h_2^2&=&9^2, &&&&&&&&&&&&\hspace{36mm}(3) \\ [1ex] s^2&+&h_1^2&=&7^2. &&&&&&&&&&&&\hspace{36mm}(4) \end{array}$ Let the brackets denote areas. We get $\begin{align*} [ABD]+[CBD]&=[ABCD] \\ \frac12(p+q+r+s)h_1+\frac12(p+q+r+s)h_2&=30 \\ \frac12(p+q+r+s)(h_1+h_2)&=30 \\ (p+q+r+s)(h_1+h_2)&=60. \hspace{49.25mm}(5) \end{align*}$ We subtract $(2)+(4)$ from $(1)+(3):$ $\begin{align*} (p+q+r)^2+(q+r+s)^2-p^2-s^2&=21 \\ \left[(p+q+r)^2-s^2\right]+\left[(q+r+s)^2-p^2\right]&=21 \\ (p+q+r+s)(p+q+r-s)+(p+q+r+s)(-p+q+r+s)&=21 \\ (p+q+r+s)(2q+2r)&=21 \\ 2(p+q+r+s)(q+r)&=21 \\ (p+q+r+s)(q+r)&=\frac{21}{2}. \hspace{9.5mm}(6) \end{align*}$ From right triangles $\triangle AEA'$ and $\triangle CEC',$ we have $\tan\theta=\frac{h_1}{r}=\frac{h_2}{q}.$ It follows that $\begin{alignat*}{8} \tan\theta&=\frac{h_1}{r}\qquad&\implies\qquad h_1&=r\tan\theta, \hspace{64mm}&(1\star)\\ \tan\theta&=\frac{h_2}{q}\qquad&\implies\qquad h_2&=q\tan\theta. &(2\star) \end{alignat*}$ Finally, we divide $(5)$ by $(6):$ $\begin{align*} \frac{h_1+h_2}{q+r}&=\frac{40}{7} \\ \frac{r\tan\theta+q\tan\theta}{q+r}&=\frac{40}{7} \hspace{15mm} &&\text{by }(1\star)\text{ and }(2\star)\\ \frac{(r+q)\tan\theta}{q+r}&=\frac{40}{7} \\ \tan\theta&=\frac{40}{7}, \end{align*}$ from which the answer is $40+7=\boxed{047}.$

~MRENTHUSIASM

Solution 3 (Bretschneider's Formula)

Bretschneider's Formula

$[asy] size(200); import olympiad; defaultpen(linewidth(0.8)+fontsize(10)); pair A,B,C,D; A=origin; B=(1.25,0); D=dir(65); C=D+0.85*dir(90)*(A-D); draw(A--B--C--D--cycle); draw(A--C^^B--D, gray+0.5); dot("$A$", A, SW); dot("$B$", B, SE); dot("$C$", C, NE); dot("$D$",D,NW); label("$a$", A--B, S); label("$b$", B--C, E); label("$c$", D--C, N); label("$d$",D--A,W); label("$u$",D--B,2*dir(170)); label("$v$",A--C,S); [/asy]$ Given quadrilateral $ABCD$ , let, $a, b, c, d$ , be the sides, $s$ the semiperimeter, and $u, v$ , the diagonals. Then the area, $K$ , is given by $K = \tfrac 14 \sqrt {4u^2v^2-(b^2+d^2-a^2-c^2)^2}$

Solution

By Bretschneider's Formula, $30=\tfrac{1}{4}\sqrt{4u^2v^2-(b^2+d^2-a^2-c^2)^2}=\tfrac{1}{4}\sqrt{4u^2v^2-441}.$ Thus, $uv=3\sqrt{1649}$ . Also, $[ABCD]=\tfrac 12 \cdot uv\sin{\theta};$ solving for $\sin{\theta}$ yields $\sin{\theta}=\tfrac{40}{\sqrt{1649}}$ . Since $\theta$ is acute, $\cos{\theta}$ is positive, from which $\cos{\theta}=\tfrac{7}{\sqrt{1649}}$ . Solving for $\tan{\theta}$ yields $\tan{\theta}=\frac{\sin{\theta}}{\cos{\theta}}=\frac{40}{7},$ for a final answer of $\boxed{047}$ .

~ Leo.Euler

Solution 4 (Symmetry)

Claim

Given an inscribed quadrilateral $ABCD$ with sides $AB = a, BC = b, CD = c,$ and $DA = d.$ Prove that the $\angle \theta < 90^\circ$ between the diagonals is given by $\begin{align*}2(ac + bd) \cos \theta = {|d^2 – c^2 + b^2 – a^2|}.\end{align*}$ Proof

Let the point $B'$ be symmetric to $B$ with respect to the perpendicular bisector $AC.$ Then the quadrilateral $AB'CD$ is an inscribed one, $AB' = b, B'C = a.$

$2 \angle AEB = \overset{\Large\frown} {AB} + \overset{\Large\frown} {CD}.$ $\begin{align*} 2\angle B'AD = \overset{\Large\frown} {B'C} + \overset{\Large\frown} {CD} = \overset{\Large\frown} {AB} + \overset{\Large\frown} {CD} \implies \angle AEB = \angle B'AD.\end{align*}$

We apply the Law of Cosines to $\triangle AB'D$ and $\triangle CB'D$ : $\begin{align*} B'D^2 = AD^2 + AB'^2 – 2 AD \cdot AB' \cos \theta, \end{align*}$ $\begin{align*} B'D^2 = CD^2 + CB'^2 + 2 CD \cdot CB' \cos \theta,\end{align*}$ $\begin{align*} d^2 + b^2 – 2 bd \cos \theta = c^2 + a^2 + 2ac \cos \theta,\end{align*}$ $\begin{align*} 2(ac + bd) \cos \theta = |d^2 – c^2 + b^2 – a^2|.\end{align*}$

vladimir.shelomovskii@gmail.com, vvsss

Note 1

By generalization, the tangent of the acute angle formed by the diagonals is $\left|\frac{4A}{a^2-b^2+c^2-d^2}\right|.$

Video Solution by MOP 2024

https://youtu.be/K0u0ACMTSw8

~r00tsOfUnity

Video Solution

https://www.youtube.com/watch?v=7DxIdTLNbo0

Video Solution by Interstigation

https://youtu.be/8GRO4za5rPI

~Interstigation

@@ Line 58: / Line 58: @@
 ^2 &= d^2 + a^2 - 2da\cos \theta. &&(4)
 \end{align*}</cmath>
-If we subtract <math>(1)+(3)</math> from <math>(2)+(4)</math>, the squared terms cancel, leaving us with:
+If we subtract <math>(2)+(4)</math> from <math>(1)+(3)</math>, the squared terms cancel, leaving us with:
 <cmath>\begin{align*}
 ^2 + 9^2 - 6^2 - 7^2 &= 2ab \cos \theta + 2bc \cos \theta + 2cd \cos \theta + 2da \cos \theta \\
@@ Line 156: / Line 156: @@
 ~ Leo.Euler
+==Solution 4 (Symmetry)==
+[[File:AIME-I-2021-12.png|350px|right]]
+<i><b>Claim</b></i>
+Given an inscribed quadrilateral <math>ABCD</math> with sides <math>AB = a, BC = b, CD = c,</math> and <math>DA = d.</math> Prove that the <math>\angle \theta < 90^\circ</math> between the diagonals is given by
+<cmath>\begin{align*}2(ac + bd) \cos \theta =  {|d^2 – c^2 + b^2 – a^2|}.\end{align*}</cmath>
+<i><b>Proof</b></i>
+Let the point <math>B'</math> be symmetric to <math>B</math> with respect to the perpendicular bisector <math>AC.</math> Then the quadrilateral <math>AB'CD</math> is an inscribed one, <math>AB' = b, B'C = a.</math>
+<cmath> 2 \angle AEB = \overset{\Large\frown} {AB}  + \overset{\Large\frown} {CD}.</cmath>
+<cmath>\begin{align*} 2\angle B'AD = \overset{\Large\frown} {B'C} + \overset{\Large\frown} {CD} = \overset{\Large\frown} {AB} + \overset{\Large\frown} {CD} \implies \angle AEB = \angle B'AD.\end{align*}</cmath>
+We apply the Law of Cosines to <math>\triangle AB'D</math> and <math>\triangle CB'D</math>:
+<cmath>\begin{align*} B'D^2 = AD^2 + AB'^2 – 2 AD \cdot AB' \cos \theta, \end{align*}</cmath>
+<cmath>\begin{align*} B'D^2 = CD^2 + CB'^2 + 2 CD \cdot CB' \cos \theta,\end{align*}</cmath>
+<cmath>\begin{align*} d^2 + b^2 – 2 bd \cos \theta = c^2 + a^2 + 2ac \cos \theta,\end{align*}</cmath>
+<cmath>\begin{align*} 2(ac + bd) \cos \theta = |d^2 – c^2 + b^2 – a^2|.\end{align*}</cmath>
+'''vladimir.shelomovskii@gmail.com, vvsss'''
+==Note 1==
+By generalization, the tangent of the acute angle formed by the diagonals is <cmath>\left|\frac{4A}{a^2-b^2+c^2-d^2}\right|.</cmath>
+==Video Solution by MOP 2024==
+https://youtu.be/K0u0ACMTSw8
+~r00tsOfUnity
 ==Video Solution==
 https://www.youtube.com/watch?v=7DxIdTLNbo0
+==Video Solution by Interstigation==
+https://youtu.be/8GRO4za5rPI
+~Interstigation
 ==See Also==
 {{AIME box|year=2021|n=II|num-b=11|num-a=13}}
 {{MAA Notice}}

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