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 $m + n$ .

Solution 1 (Sines and Cosines)

Solution 1.1 (Version 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 1.2 (Version 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: $\begin{align*} 5^2 + 9^2 - 6^2 - 7^2 &= 2ab \cos \theta + 2bc \cos \theta + 2cd \cos \theta + 2da \cos \theta \\ 21 &= 2\cos \theta (ab + bc + cd + da). \end{align*}$ 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 $40 + 7 = \boxed{047}$ .

~ my_aops_lessons

Solution 2 (Right Triangles)

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:

File: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

2021 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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