Problem: Convex pentagon $ABCDE$ has side lengths $AB=5$ $BC=CD=DE=6$ , and $EA=7$ . Moreover, the pentagon has an inscribed circle (a circle tangent to each side of the pentagon). Find the area of $ABCDE$

Solution: Let $\omega$ be the inscribed circle, $I$ be its center, and $r$ be its radius. The area of $ABCDE$ is equal to its semiperimeter, $15,$ times $r$ , so the problem is reduced to finding $r$ . Let $a$ be the length of the tangent segment from $A$ to $\omega$ , and analogously define $b$ $c$ $d$ , and $e$ . Then $a+b=5$ $b+c= c+d=d+e=6$ , and $e+a=7$ , with a total of $a+b+c+d+e=15$ . Hence $a=3$ $b=d=2$ , and $c=e=4$ . It follows that $\angle B= \angle D$ and $\angle C= \angle E$ . Let $Q$ be the point where $\omega$ is tangent to $\overline{CD}$ . Then $\angle IAE = \angle IAB =\frac{1}{2}\angle A$ . Now we claim that points $A, I, Q$ are collinear, which can be proved if $\angle{AIQ}=\angle{QIA}=180^{\circ}$ . The sum of the internal angles in polygons $ABCQI$ and $AIQDE$ are equal, so $\angle IAE + \angle AIQ + \angle IQD + \angle D + \angle E = \angle IAB + \angle B + \angle C + \angle CQI + \angle QIA$ , which implies that $\angle AIQ$ must be $180^\circ$ . Therefore points $A$ $I$ , and $Q$ are collinear. [asy] defaultpen(fontsize(8pt)); unitsize(0.025cm);  pair[] vertices = {(0,0), (5,0), (8.6,4.8), (3.8,8.4), (-1.96, 6.72)}; string[] labels = {"$A$", "$B$", "$C$", "$D$", "$E$"}; pair[] dirs = {SW, SE,E, N, NW};   string[] smallLabels = {"$a$", "$b$", "$c$",  "$d$", "$e$"};  pair I = (3,4);  real rad = 4; pair Q = foot(I, vertices[2], vertices[3]);  pair[] interpoints = {};  for(int i =0; i<vertices.length; ++i){    interpoints.push(foot(I, vertices[i], vertices[(i+1)%vertices.length])); }    for(int i = 0; i< vertices.length; ++i){    draw(vertices[i]--vertices[(i+1)%vertices.length]);    dot(labels[i],vertices[i],dirs[i]);    draw(I--vertices[i]); } draw(Circle(I, rad)); dot("$I$", I, dir(200));  draw(I--Q);  dot("$Q$", Q, NE);  for(int i = 0; i < vertices.length; ++i){     label(smallLabels[i], vertices[i] --interpoints[i]);     //dot(interpoints[i], blue);    label(smallLabels[i], interpoints[(i-1)%vertices.length] -- vertices[i]); }  [/asy] Because $\overline{AQ} \perp \overline{CD}$ , it follows that \[AC^2-AD^2=CQ^2-DQ^2=c^2-d^2=12.\] Another expression for $AC^2-AD^2$ can be found as follows. Note that $\tan \left(\frac{\angle B}{2}\right) = \frac{r}{2}$ and $\tan \left(\frac{\angle E}{2}\right) = \frac{r}{4}$ , so \[\cos (\angle B) =\frac{1-\tan^2 \left(\frac{\angle B}{2}\right)}{1+\tan^2 \left(\frac{\angle B}{2}\right)} =  \frac{4-r^2}{4+r^2}\] and \[\cos (\angle E) = \frac{1-\tan^2 \left(\frac{\angle E}{2}\right)}{1+\tan^2 \left(\frac{\angle E}{2}\right)}= \frac{16-r^2}{16+r^2}.\] Applying the Law of Cosines to $\triangle ABC$ and $\triangle AED$ gives \[AC^2=AB^2+BC^2-2\cdot AB\cdot BC\cdot \cos (\angle B) = 5^2+6^2-2 \cdot 5 \cdot 6 \cdot \frac{4-r^2}{4+r^2}\] and \[AD^2=AE^2+DE^2-2 \cdot AE \cdot DE \cdot \cos(\angle E) = 7^2+6^2-2 \cdot 7 \cdot 6 \cdot \frac{16-r^2}{16+r^2}.\] Hence
\[12=AC^2- AD^2= 5^2-2\cdot 5 \cdot 6\cdot \frac{4-r^2}{4+r^2} -7^2+2\cdot 7 \cdot 6 \cdot \frac{16-r^2}{16+r^2},\] yielding \[2\cdot 7 \cdot 6 \cdot \frac{16-r^2}{16+r^2}- 2\cdot 5 \cdot 6\cdot \frac{4-r^2}{4+r^2}= 36;\] equivalently \[7(16-r^2)(4+r^2)-5(4-r^2)(16+r^2) = 3(16+r^2)(4+r^2).\] Substituting $x=r^2$ gives the quadratic equation $5x^2-84x+64=0$ , with solutions $\frac{42 - 38}{5}=\frac45$ , and $\frac{42 + 38}{5}= 16$ . The solution $r^2=\frac45$ corresponds to a five-pointed star, which is not convex. Indeed, if $r<3$ , then $\tan \left(\frac{\angle A}{2}\right)$ $\tan \left(\frac{\angle C}{2}\right)$ , and $\tan \left(\frac{\angle E}{2}\right)$ are less than $1,$ implying that $\angle A$ $\angle C$ , and $\angle E$ are acute, which cannot happen in a convex pentagon. Thus $r^2=16$ and $r=4$ . The requested area is $15\cdot4 = \boxed{60}$

Answer:  60