# 2021 AIME II Problems/Problem 12

## 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 $m + n$.

## Diagram

~MRENTHUSIASM (by Geometry Expressions)

## 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 $\theta = \angle AED$. Hence, $\angle AEB = \angle CED = 180^\circ - \theta$ and $\angle BEC = \theta$.

First, we use the triangle area formula with sines to write down an equation of the area of the quadrilateral $ABCD$.

We have \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*} 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 $\sin \left( 180^\circ - \theta \right) = \sin \theta$.

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

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

In $\triangle AEB$, following from the law of cosines, we have $$a^2 + b^2 - 2 a b \cos \angle AEB = AB^2 .$$

Because $\cos \angle AEB = \cos \left( 180^\circ - \theta \right) = \cos \theta$ and $AB = 5$, we have $$a^2 + b^2 + 2 a b \cos \theta = 5^2 . \ \ \ (2)$$

In $\triangle BEC$, following from the law of cosines, we have $$b^2 + c^2 - 2 b c \cos \angle BEC = BC^2 .$$

Because $\cos \angle AEB = \cos \theta$ and $BC = 6$, we have $$b^2 + c^2 - 2 b c \cos \theta = 6^2 . \ \ \ (3)$$

In $\triangle CED$, following from the law of cosines, we have $$c^2 + d^2 - 2 c d \cos \angle CED = CD^2 .$$

Because $\cos \angle CED = \cos \left( 180^\circ - \theta \right) = \cos \theta$ and $CD = 9$, we have $$c^2 + d^2 + 2 c d \cos \theta = 9^2 . \ \ \ (4)$$

In $\triangle DEA$, following from the law of cosines, we have $$d^2 + a^2 - 2 d a \cos \angle DEA = DA^2 .$$

Because $\cos \angle DEC = \cos \theta$ and $DA = 7$, we have $$d^2 + a^2 - 2 d a \cos \theta = 7^2 . \ \ \ (5)$$

By taking $\frac{1}{2} \left( {\rm Eq} \ (2) - {\rm Eq} \ (3) + {\rm Eq} \ (4) - {\rm Eq} \ (5) \right)$, we get

$$\left( ab + bc + cd + da \right) \cos \theta = \frac{21}{2} . \ \ \ (6)$$

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

~ 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

## Solution 3 (Pythagorean Theorem and Right Triangle Trigonometry)

This solution refers to the Diagram section.

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:

WILL BE BACK SOON.

~MRENTHUSIASM