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

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(Solution 1 (Sines and Cosines))
 
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==Problem==
 
==Problem==
These problems will not be posted until the 2021 AIME II is released on Thursday, March 25, 2021.
+
A convex quadrilateral has area <math>30</math> and side lengths <math>5, 6, 9,</math> and <math>7,</math> in that order. Denote by <math>\theta</math> the measure of the acute angle formed by the diagonals of the quadrilateral. Then <math>\tan \theta</math> can be written in the form <math>\tfrac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>.
==Solution==
 
We can't have a solution without a problem.
 
  
==See also==
+
==Solution 1 (Sines and Cosines)==
 +
Since we are asked to find <math>\tan \theta</math>, we can find <math>\sin \theta</math> and <math>\cos \theta</math> separately and use their values to get <math>\tan \theta</math>. We can start by drawing a diagram. Let the vertices of the quadrilateral be <math>A</math>, <math>B</math>, <math>C</math>, and <math>D</math>. Let <math>AB = 5</math>, <math>BC = 6</math>, <math>CD = 9</math>, and <math>DA = 7</math>. Let <math>AX = a</math>, <math>BX = b</math>, <math>CX = c</math>, and <math>DX = d</math>. We know that <math>\theta</math> is the acute angle formed between the intersection of the diagonals <math>AC</math> and <math>BD</math>.
 +
<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);
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X = intersectionpoint(A--C,B--D);
 +
draw(A--C);
 +
draw(B--D);
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label("$A$",A,NW);
 +
label("$B$",B,NE);
 +
label("$C$",C,SE);
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label("$D$",D,SW);
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dot(X);
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label("$X$",X,S);
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label("$5$",(A+B)/2,N);
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label("$6$",(B+C)/2,E);
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label("$9$",(C+D)/2,S);
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label("$7$",(D+A)/2,W);
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label("$\theta$",X,2.5E);
 +
label("$a$",(A+X)/2,NE);
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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 <math>ABCD</math> is <math>30</math>. We can express this area using the areas of triangles <math>AXB</math>, <math>BXC</math>, <math>CXD</math>, and <math>DXA</math>. Since we want to find <math>\sin \theta</math> and <math>\cos \theta</math>, we can represent these areas using <math>\sin \theta</math> as follows:
 +
<cmath>\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*}</cmath>
 +
We know that <math>\sin (180^\circ - \theta) = \sin \theta</math>. Therefore it follows that:
 +
<cmath>\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*}</cmath>
 +
From here we see that <math>\sin \theta = \frac{60}{ab + bc + cd + da}</math>. Now we need to find <math>\cos \theta</math>. Using the Law of Cosines on each of the four smaller triangles, we get following equations:
 +
<cmath>\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*}</cmath>
 +
We know that <math>\cos (180^\circ - \theta) = -\cos \theta</math> for all <math>\theta</math>. We can substitute this value into our equations to get:
 +
<cmath>\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*}</cmath>
 +
If we subtract <math>(1)+(3)</math> from <math>(2)+(4)</math>, the squared terms cancel, leaving us with:
 +
<cmath>\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*}</cmath>
 +
From here we see that <math>\cos \theta = \frac{21/2}{ab + bc + cd + da}</math>.
 +
 
 +
Since we have figured out <math>\sin \theta</math> and <math>\cos \theta</math>, we can calculate <math>\tan \theta</math>: <cmath>\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}.</cmath>
 +
Therefore our answer is <math>40 + 7 = \boxed{047}</math>.
 +
 
 +
~ Steven Chen (www.professorchenedu.com)
 +
 
 +
~ my_aops_lessons
 +
 
 +
==Solution 2 (Right Triangles)==
 +
In convex quadrilateral <math>ABCD,</math> let <math>AB=5,BC=6,CD=9,</math> and <math>DA=7.</math> Let <math>A'</math> and <math>C'</math> be the feet of the perpendiculars from <math>A</math> and <math>C,</math> respectively, to <math>\overline{BD}.</math> We obtain the following diagram:
 +
[[File:2021 AIME II Problem 12 Solution.png|center]]
 +
Let <math>BC'=p,C'E=q,EA'=r,A'D=s,AA'=h_1,</math> and <math>CC'=h_2.</math> We apply the Pythagorean Theorem to right triangles <math>\triangle ABA',\triangle BCC',\triangle CDC',</math> and <math>\triangle DAA',</math> respectively:
 +
<cmath>\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}</cmath>
 +
Let the brackets denote areas. We get
 +
<cmath>\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*}</cmath>
 +
We subtract <math>(2)+(4)</math> from <math>(1)+(3):</math>
 +
<cmath>\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*}</cmath>
 +
From right triangles <math>\triangle AEA'</math> and <math>\triangle CEC',</math> we have <math>\tan\theta=\frac{h_1}{r}=\frac{h_2}{q}.</math> It follows that
 +
<cmath>\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*}</cmath>
 +
Finally, we divide <math>(5)</math> by <math>(6):</math>
 +
<cmath>\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*}</cmath>
 +
from which the answer is <math>40+7=\boxed{047}.</math>
 +
 
 +
~MRENTHUSIASM
 +
 
 +
==Solution 3==
 +
Let <math>A</math>, <math>B</math>, <math>C</math>, <math>D</math> be the vertices of the quadrilateral, <math>a</math>, <math>b</math>, <math>c</math>, <math>d</math> be the lengths of the sides of <math>ABCD</math>, and <math>p</math> and <math>q</math> be the lengths of the diagonals of <math>ABCD</math>. By Bretschneider's formula, <math>30=\frac{1}{4} \cdot \sqrt{4p^2q^2-(b^2+d^2-a^2-c^2)^2}=\frac{1}{4} \cdot \sqrt{4p^2q^2-441}</math>. Thus, <math>pq=\sqrt{14841}</math>. Also, <math>[ABCD]=\frac{pq \cdot \sin{\theta}}{2}</math>; solving for <math>\sin{\theta}</math> yields <math>\sin{\theta}=\frac{120}{\sqrt{14841}}</math>. Since <math>\theta</math> is acute, <math>\cos{\theta}</math> is positive, so <math>\cos{\theta}=\frac{21}{\sqrt{14841}}</math>. Solving for <math>\tan{\theta}</math> yields <math>\tan{\theta}=\sin{\theta}/\cos{\theta}=\frac{40}{7}</math>, for a final answer of <math>\boxed{047}</math>.
 +
 
 +
~ Leo.Euler
 +
 
 +
==Video Solution==
 +
https://www.youtube.com/watch?v=7DxIdTLNbo0
 +
 
 +
==See Also==
 
{{AIME box|year=2021|n=II|num-b=11|num-a=13}}
 
{{AIME box|year=2021|n=II|num-b=11|num-a=13}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 00:14, 7 September 2021

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 $(1)+(3)$ from $(2)+(4)$, 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:

2021 AIME II Problem 12 Solution.png

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

Let $A$, $B$, $C$, $D$ be the vertices of the quadrilateral, $a$, $b$, $c$, $d$ be the lengths of the sides of $ABCD$, and $p$ and $q$ be the lengths of the diagonals of $ABCD$. By Bretschneider's formula, $30=\frac{1}{4} \cdot \sqrt{4p^2q^2-(b^2+d^2-a^2-c^2)^2}=\frac{1}{4} \cdot \sqrt{4p^2q^2-441}$. Thus, $pq=\sqrt{14841}$. Also, $[ABCD]=\frac{pq \cdot \sin{\theta}}{2}$; solving for $\sin{\theta}$ yields $\sin{\theta}=\frac{120}{\sqrt{14841}}$. Since $\theta$ is acute, $\cos{\theta}$ is positive, so $\cos{\theta}=\frac{21}{\sqrt{14841}}$. Solving for $\tan{\theta}$ yields $\tan{\theta}=\sin{\theta}/\cos{\theta}=\frac{40}{7}$, for a final answer of $\boxed{047}$.

~ Leo.Euler

Video Solution

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

See Also

2021 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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