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: 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 $(2)+(4)$ from $(1)+(3)$ , 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 $40 + 7 = \boxed{047}$

Answer:  047