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

m (Solution 3 (Pythagorean Theorem and Right Triangle Trigonometry))
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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>.
 
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>.
  
==Diagram==
+
==Solution 1 (Sines and Cosines)==
[[File:2021 AIME II Problem 12 Diagram.png|center]]
 
 
 
~MRENTHUSIASM (by Geometry Expressions)
 
 
 
==Solution 1==
 
 
 
We denote by <math>A</math>, <math>B</math>, <math>C</math> and <math>D</math> four vertices of this quadrilateral, such that <math>AB = 5</math>, <math>BC = 6</math>, <math>CD = 9</math>, <math>DA = 7</math>.
 
We denote by <math>E</math> the point that two diagonals <math>AC</math> and <math>BD</math> meet at.
 
To simplify the notation, we denote <math>a = AE</math>, <math>b = BE</math>, <math>c = CE</math>, <math>d = DE</math>.
 
 
 
We denote <math>\theta = \angle AED</math>.
 
Hence, <math>\angle AEB = \angle CED = 180^\circ - \theta</math> and <math>\angle BEC = \theta</math>.
 
 
 
First, we use the triangle area formula with sines to write down an equation of the area of the quadrilateral <math>ABCD</math>.
 
 
 
We have
 
<cmath>
 
\begin{align*}
 
{\rm Area} \ ABCD
 
& = {\rm Area} \ \triangle ABE + {\rm Area} \ \triangle BCE + {\rm Area} \ \triangle CDE + {\rm Area} \ \triangle DAE \\
 
& = \frac{1}{2} ab \sin \angle AEB + \frac{1}{2} bc \sin \angle BEC + \frac{1}{2} cd \sin \angle CED + \frac{1}{2} da \sin \angle DEA \\
 
& = \frac{1}{2} ab \sin \theta + \frac{1}{2} bc \sin \left( 180^\circ - \theta \right)
 
+ \frac{1}{2} cd \sin \theta + \frac{1}{2} da \sin \left( 180^\circ - \theta \right) \\
 
& = \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} \left( ab + bc + cd + da \right) \sin \theta ,
 
\end{align*}
 
</cmath>
 
where the second equality follows from the formula to use the sine function to compute a triangle area, the the fourth equality follows from the property that <math>\sin \left( 180^\circ - \theta \right) = \sin \theta</math>.
 
 
 
Because <math>{\rm Area} \ ABCD = 30</math>, we have
 
<cmath>
 
\[
 
\left( ab + bc + cd + da \right) \sin \theta = 60 . \ \ \ (1)
 
\]
 
</cmath>.
 
 
 
Second, we use the law of cosines to establish four equations for four sides of the quadrilateral <math>ABCD</math>.
 
 
 
In <math>\triangle AEB</math>, following from the law of cosines, we have
 
<cmath>
 
\[
 
a^2 + b^2 - 2 a b \cos \angle AEB
 
=  AB^2 .
 
\]
 
</cmath>
 
 
 
Because <math>\cos \angle AEB = \cos \left( 180^\circ - \theta \right) = \cos \theta</math> and <math>AB = 5</math>, we have
 
<cmath>
 
\[
 
a^2 + b^2 + 2 a b \cos \theta  = 5^2 . \ \ \ (2)
 
\]
 
</cmath>
 
 
 
In <math>\triangle BEC</math>, following from the law of cosines, we have
 
<cmath>
 
\[
 
b^2 + c^2 - 2 b c \cos \angle BEC
 
=  BC^2 .
 
\]
 
</cmath>
 
 
 
Because <math>\cos \angle AEB = \cos \theta</math> and <math>BC = 6</math>, we have
 
<cmath>
 
\[
 
b^2 + c^2 - 2 b c \cos \theta  = 6^2 . \ \ \ (3)
 
\]
 
</cmath>
 
 
 
In <math>\triangle CED</math>, following from the law of cosines, we have
 
<cmath>
 
\[
 
c^2 + d^2 - 2  c d \cos \angle CED
 
=  CD^2 .
 
\]
 
</cmath>
 
 
 
Because <math>\cos \angle CED = \cos \left( 180^\circ - \theta \right) = \cos \theta</math> and <math>CD = 9</math>, we have
 
<cmath>
 
\[
 
c^2 + d^2 + 2 c d \cos \theta  = 9^2 . \ \ \ (4)
 
\]
 
</cmath>
 
 
 
In <math>\triangle DEA</math>, following from the law of cosines, we have
 
<cmath>
 
\[
 
d^2 + a^2 - 2  d a \cos \angle DEA
 
=  DA^2 .
 
\]
 
</cmath>
 
 
 
Because <math>\cos \angle DEC = \cos \theta</math> and <math>DA = 7</math>, we have
 
<cmath>
 
\[
 
d^2 + a^2 - 2 d a \cos \theta  = 7^2 . \ \ \ (5)
 
\]
 
</cmath>
 
 
 
By taking <math>\frac{1}{2} \left( {\rm Eq} \ (2) - {\rm Eq} \ (3) + {\rm Eq} \ (4) - {\rm Eq} \ (5) \right)</math>, we get
 
 
 
<cmath>
 
\[
 
\left( ab + bc + cd + da \right) \cos \theta = \frac{21}{2} . \ \ \ (6)
 
\]
 
</cmath>
 
 
 
By taking <math>\frac{{\rm Eq} \ (1)}{{\rm Eq} \ (6)}</math>, we get
 
<cmath>
 
\[
 
\tan \theta = \frac{60}{21/2} = \frac{40}{7} .
 
\]
 
</cmath>
 
 
 
Therefore, by writing this answer in the form of <math>\frac{m}{n}</math>, we have <math>m = 40</math> and <math>n = 7</math>.
 
Therefore, the answer to this question is <math>m + n = 40 + 7 = \boxed{047}</math>.
 
 
 
~ Steven Chen (www.professorchenedu.com)
 
 
 
==Solution 2==
 
 
 
 
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>.
 
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>
 
<asy>
 
unitsize(4cm);
 
unitsize(4cm);
Line 153: Line 31:
 
label("$d$",(D+X)/2,SE);
 
label("$d$",(D+X)/2,SE);
 
</asy>
 
</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:
 
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*}
 
<cmath>\begin{align*}
 
30 &=[ABCD] \\
 
30 &=[ABCD] \\
 
&=[AXB] + [BXC] + [CXD] + [DXA] \\
 
&=[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 (\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)
+
&=\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>
 
\end{align*}</cmath>
 
 
We know that <math>\sin (180^\circ - \theta) = \sin \theta</math>. Therefore it follows that:
 
We know that <math>\sin (180^\circ - \theta) = \sin \theta</math>. Therefore it follows that:
 
 
<cmath>\begin{align*}
 
<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) \\
 
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} 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)
+
&=\frac{1}{2}\sin\theta (ab + bc + cd + da).
 
\end{align*}</cmath>
 
\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:
 
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*}   
 
<cmath>\begin{align*}   
5^2 &= a^2 + b^2 - 2ab\cos(180^\circ-\theta) \\   
+
5^2 &= a^2 + b^2 - 2ab\cos(180^\circ-\theta), \\   
6^2 &= b^2 + c^2 - 2bc\cos \theta \\   
+
6^2 &= b^2 + c^2 - 2bc\cos \theta, \\   
9^2 &= c^2 + d^2 - 2cd\cos(180^\circ-\theta) \\   
+
9^2 &= c^2 + d^2 - 2cd\cos(180^\circ-\theta), \\   
7^2 &= d^2 + a^2 - 2da\cos \theta   
+
7^2 &= d^2 + a^2 - 2da\cos \theta.  
 
\end{align*}</cmath>
 
\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:
We know that <math>\cos (180^\circ - \theta) = -\cos \theta</math>. We can substitute this value into our equations to get:
 
 
 
 
<cmath>\begin{align*}   
 
<cmath>\begin{align*}   
5^2 &= a^2 + b^2 + 2ab\cos \theta \\   
+
5^2 &= a^2 + b^2 + 2ab\cos \theta, &&(1) \\   
6^2 &= b^2 + c^2 - 2bc\cos \theta \\   
+
6^2 &= b^2 + c^2 - 2bc\cos \theta, &&(2) \\   
9^2 &= c^2 + d^2 + 2cd\cos \theta \\   
+
9^2 &= c^2 + d^2 + 2cd\cos \theta, &&(3) \\   
7^2 &= d^2 + a^2 - 2da\cos \theta   
+
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>
 
\end{align*}</cmath>
 
If we subtract the sum of the first and third equation from the sum of the second and fourth equation, the squared terms cancel, leaving us with:
 
<cmath>5^2 + 9^2 - 6^2 - 7^2 = 2ab \cos \theta + 2bc \cos \theta + 2cd \cos \theta + 2da \cos \theta</cmath>
 
<cmath>21 = 2\cos \theta (ab  + bc + cd + da)</cmath>
 
 
 
From here we see that <math>\cos \theta = \frac{21/2}{ab + bc + cd + da}</math>.
 
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>:
+
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>.
  
<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>
+
~ Steven Chen (www.professorchenedu.com)
 
 
Therefore our answer is <math>40 + 7 = \boxed{047}</math>.
 
  
 
~ my_aops_lessons
 
~ my_aops_lessons
  
==Solution 3 (Pythagorean Theorem and Right Triangle Trigonometry)==
+
==Solution 2 (Right Triangles)==
This solution refers to the <b>Diagram</b> section.
+
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:
 +
<asy>
 +
/* Made by MRENTHUSIASM */
 +
size(500);
  
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:
+
pair A, B, C, D, P, A1, C1;
[[File:2021 AIME II Problem 12 Solution.png|center]]
+
B = origin;
Let <math>BC'=p,C'E=q,EA'=r,A'D=s,AA'=h_1,</math> and <math>CC'=h_2.</math> Applying the Pythagorean Theorem to right triangles <math>\triangle ABA',\triangle BCC',\triangle CDC',</math> and <math>\triangle DAA',</math> we respectively get
+
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 <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}
 
<cmath>\begin{array}{ccccccccccccccccc}
 
(p+q+r)^2&+&h_1^2&=&5^2, &&&&&&&&&&&&\hspace{36mm}(1) \\ [1ex]
 
(p+q+r)^2&+&h_1^2&=&5^2, &&&&&&&&&&&&\hspace{36mm}(1) \\ [1ex]
Line 239: Line 131:
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
 
\frac{h_1+h_2}{q+r}&=\frac{40}{7} \\
 
\frac{h_1+h_2}{q+r}&=\frac{40}{7} \\
\frac{\overbrace{r\tan\theta}^{\text{by }(1\star).}+\overbrace{q\tan\theta}^{\text{by }(2\star).}}{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} \\
 
\frac{(r+q)\tan\theta}{q+r}&=\frac{40}{7} \\
 
\tan\theta&=\frac{40}{7},
 
\tan\theta&=\frac{40}{7},
Line 246: Line 138:
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM
 +
 +
==Solution 3 (Bretschneider's Formula)==
 +
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}\sqrt{4p^2q^2-(b^2+d^2-a^2-c^2)^2}=\frac{1}{4}\sqrt{4p^2q^2-441}</math>. Thus, <math>pq=3\sqrt{1649}</math>. Also, <math>[ABCD]=\frac{pq\sin{\theta}}{2}</math>; solving for <math>\sin{\theta}</math> yields <math>\sin{\theta}=\frac{40}{\sqrt{1649}}</math>. Since <math>\theta</math> is acute, <math>\cos{\theta}</math> is positive, from which <math>\cos{\theta}=\frac{7}{\sqrt{1649}}</math>. Solving for <math>\tan{\theta}</math> yields <math>\tan{\theta}=\frac{\sin{\theta}}{\cos{\theta}}=\frac{40}{7}</math>, for a final answer of <math>\boxed{047}</math>.
 +
 +
~ Leo.Euler
  
 
==Video Solution==
 
==Video Solution==
 
https://www.youtube.com/watch?v=7DxIdTLNbo0
 
https://www.youtube.com/watch?v=7DxIdTLNbo0
  
==See also==
+
==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 21:33, 2 October 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: [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)

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, $30=\frac{1}{4}\sqrt{4p^2q^2-(b^2+d^2-a^2-c^2)^2}=\frac{1}{4}\sqrt{4p^2q^2-441}$. Thus, $pq=3\sqrt{1649}$. Also, $[ABCD]=\frac{pq\sin{\theta}}{2}$; solving for $\sin{\theta}$ yields $\sin{\theta}=\frac{40}{\sqrt{1649}}$. Since $\theta$ is acute, $\cos{\theta}$ is positive, from which $\cos{\theta}=\frac{7}{\sqrt{1649}}$. Solving for $\tan{\theta}$ yields $\tan{\theta}=\frac{\sin{\theta}}{\cos{\theta}}=\frac{40}{7}$, for a final answer of $\boxed{047}$.

~ Leo.Euler

Video Solution

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

See Also

2021 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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All AIME Problems and Solutions

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