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

(Solution 1)
(Solution 1)
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~ Steven Chen (www.professorchenedu.com)
 
~ Steven Chen (www.professorchenedu.com)
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 +
==Solution 2==
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 +
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>.
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 +
<asy>
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unitsize(4cm);
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pair A,B,C,D,X;
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A = (0,0);
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B = (1,0);
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C = (1.25,-1);
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D = (-0.75,-0.75);
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draw(A--B--C--D--cycle,black+1bp);
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X = intersectionpoint(A--C,B--D);
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draw(A--C);
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draw(B--D);
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label("$A$",A,NW);
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label("$B$",B,NE);
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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);
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label("$a$",(A+X)/2,NE);
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label("$b$",(B+X)/2,NW);
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label("$c$",(C+X)/2,SW);
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label("$d$",(D+X)/2,SE);
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</asy>
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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:
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 +
<cmath>\begin{align*}
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30 &=[ABCD] \\
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&=[AXB] + [BXC] + [CXD] + [DXA] \\
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&=\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) \\
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&=\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)
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\end{align*}</cmath>
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We know that <math>\sin (180^\circ - \theta) = \sin \theta</math>. Therefore it follows that:
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 +
<cmath>\begin{align*}
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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) \\
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&=\frac{1}{2} ab \sin (\theta) + \frac{1}{2} bc \sin (\theta) + \frac{1}{2} cd \sin (\theta) + \frac{1}{2} da \sin (\theta) \\
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&=\frac{1}{2}\sin\theta (ab + bc + cd + da)
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\end{align*}</cmath>
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 +
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:
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 +
<cmath>\begin{align*} 
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5^2 &= a^2 + b^2 - 2ab\cos(180^\circ-\theta) \\ 
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6^2 &= b^2 + c^2 - 2bc\cos \theta \\ 
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9^2 &= c^2 + d^2 - 2cd\cos(180^\circ-\theta) \\ 
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7^2 &= d^2 + a^2 - 2da\cos \theta 
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\end{align*}</cmath>
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 +
We know that <math>\cos (180^\circ - \theta) = -\cos \theta</math>. We can substitute this value into our equations to get:
 +
 +
<cmath>\begin{align*} 
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5^2 &= a^2 + b^2 + 2ab\cos \theta \\ 
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6^2 &= b^2 + c^2 - 2bc\cos \theta \\ 
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9^2 &= c^2 + d^2 + 2cd\cos \theta \\ 
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7^2 &= d^2 + a^2 - 2da\cos \theta 
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\end{align*}</cmath>
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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:
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<cmath>5^2 + 9^2 - 6^2 - 7^2 = 2ab \cos \theta + 2bc \cos \theta + 2cd \cos \theta + 2da \cos \theta</cmath>
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<cmath>21 = 2\cos \theta (ab  + bc + cd + da)</cmath>
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 +
From here we see that <math>\cos \theta = \frac{21/2}{ab + bc + cd + da}</math>.
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 +
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>
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Therefore our answer is <math>40 + 7 = \boxed{047}</math>.
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 +
~ my_aops_lessons
  
 
==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}}

Revision as of 18:01, 22 March 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

We denote by $A$, $B$, $C$ and $D$ four vertices of this quadrilateral, such that $AB = 5$, $BC = 6$, $CD = 9$, $DA = 7$. We denote by $E$ the point that two diagonals $AC$ and $BD$ meet at. To simplify the notation, we denote $a = AE$, $b = BE$, $c = CE$, $d = DE$. We denote $\theta = \angle AED$.

First, we use the triangle area formula with sines to write down an equation of the area of the quadrilateral $ABCD$. We have ${\rm Area} \ ABCD = {\rm Area} \ \triangle ABE + {\rm Area} \ \triangle BCE + {\rm Area} \ \triangle CDE + {\rm Area} \ \triangle DAE = \frac{1}{2} ab \sin \angle AEB + \frac{1}{2} bc \sin \angle BEC + \frac{1}{2} cd \sin \angle CED + \frac{1}{2} da \sin \angle DEA = \frac{1}{2} \left( ab + bc + cd + da \right) \sin \theta$.

Because ${\rm Area} \ ABCD = 30$, we have $\left( ab + bc + cd + da \right) \sin \theta = 60$. We index this equation as Eq (1).

Second, we use the law of cosines to establish four equations for four sides of the quadrilateral $ABCD$.

By applying the law of cosines to $\triangle AEB$, we have $a^2 + b^2 - 2 a b \cos \angle AEB = AB^2 = 5^2$. Note that $\cos \angle AEB = \cos \left( 180^\circ - \theta \right) = \cos \theta$.

Hence, $a^2 + b^2 + 2 a b \cos \theta = 5^2$. We index this equation as Eq (2).

Analogously, we can establish the following equation for $\triangle BEC$ that $b^2 + c^2 - 2 b c \cos \theta = 6^2$ (indexed as Eq (3)),

the following equation for $\triangle CED$ that $c^2 + d^2 + 2 c d \cos \theta = 9^2$ (indexed as Eq (4)),

and the following equation for $\triangle DEA$ that $d^2 + a^2 - 2 d a \cos \theta = 7^2$ (indexed as Eq (5)).

By taking Eq (2) - Eq (3) + Eq (4) - Eq (5) and dividing both sides of the equation by 2, we get $\left( ab + bc + cd + da \right) \cos \theta = \frac{21}{2}$. We index this equation as Eq (6).

By taking $\frac{{\rm Eq} \ (1)}{{\rm Eq} \ (6)}$, we get $\tan \theta = \frac{60}{21/2} = \frac{40}{7}$.

Therefore, by writing this answer in the form of $\frac{m}{n}$, we have $m = 40$ and $n = 7$. Therefore, the answer to this question is $m + n = 40 + 7 = 47$.

~ Steven Chen (www.professorchenedu.com)

Solution 2

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 $\cos (180^\circ - \theta) = -\cos \theta$. We can substitute this value into our equations to get:

\begin{align*}   5^2 &= a^2 + b^2 + 2ab\cos \theta \\   6^2 &= b^2 + c^2 - 2bc\cos \theta \\   9^2 &= c^2 + d^2 + 2cd\cos \theta \\   7^2 &= d^2 + a^2 - 2da\cos \theta   \end{align*}

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: \[5^2 + 9^2 - 6^2 - 7^2 = 2ab \cos \theta + 2bc \cos \theta + 2cd \cos \theta + 2da \cos \theta\] \[21 = 2\cos \theta (ab  + bc + cd + da)\]

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 $40 + 7 = \boxed{047}$.

~ my_aops_lessons

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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