Three Incircles and a Rectangle

Suppose that $AD=a$ and $AB=b$ , where $a > b$ are positive integers. Then $AF=BE=\sqrt{a^2-b^2}$ , $DF=DC=b$ and $FE=CE=a-\sqrt{a^2-b^2}$ . Thus the quadrilateral $CDFE$ has an incircle with radius $r$ , where $\tfrac12\big(b + a - \sqrt{a^2-b^2}\big)r \; = \; \tfrac12b\big(a-\sqrt{a^2-b^2}\big)$ Since $r=\tfrac34$ we deduce that $\begin{aligned} 3\big(a + b - \sqrt{a^2-b^2}\big) & = \; 4b\big(a - \sqrt{a^2-b^2}\big) \\ (4b-3)\sqrt{a^2-b^2} & = \; 4ab - 3a-3b \\ (4b-3)^2(a^2-b^2) & = \; (4ab - 3a - 3b)^2 \\ 8b^3 - 12b^2 -12ab + 9a+9b & = \; 0 \end{aligned}$ From this it follows that $b = 3c$ is divisible by $3$ , and hence that $\begin{aligned}0 & = \; 24c^3 - 12c^2 - 4ac + 3c + a \\ a & = \; \frac{24c^3 - 12c^2 + 3c}{4c-1} \; = \; 6c^2 - \frac{3c(2c-1)}{4c-1} \end{aligned}$ and hence $4c-1$ must divide $3c(2c-1)$ . SInce $4 \times c - (4c-1) = 1$ , $c$ and $4c-1$ are coprime. Since $(4c-1) - 2(2c-1) = 1$ , $2c-1$ and $4c-1$ are coprime. Thus we deduce that $4c-1$ must divide $3$ , and hence that $c=1$ .

Thus the only solution is given by $a=5$ , $b=3$ , and so the area of the rectangle is $\boxed{15}$ .

Let $AB=CD=a$ and $BC=DA=b$ . We note that $\triangle ADF$ and $\triangle ABE$ are congruent. Let the center of the incircle of $FECD$ be $O$ , $OG=\frac 34$ and $OH=\frac 34$ be perpendicular to $CD$ and $BC$ respectively, and $\angle CDF = \angle DAF = \angle AEB = \theta$ . Then we have:

$\begin{aligned} CD & = DG + GC \\ a & = \frac 34 \cot \frac \theta 2 + \frac 34 & \small \blue{\text{Let }t = \tan \frac \theta 2} \\ a & = \frac 34 \left(\frac 1t + 1\right) \\ \implies t & = \frac 3{4a-3} \end{aligned}$

From the congruent triangles,

$\begin{aligned} b & = a \csc \theta = a \cdot \frac {1+t^2}{2t} \\ & = a \left(\frac 1t + t \right) = \frac a2 \left(\frac {4a-3}3 + \frac 3{4a-3} \right) \\ \implies 6b & = 4a^2 - 3a + \frac {9a}{4a-3} \end{aligned}$

Putting in integer values for $a$ it is found that the smallest integer solution is when $a=3$ , then $b=5$ and the area of the rectangle is $ab = \boxed {15}$ .

Since $\triangle ABE \sim \triangle DFA$ by AA similarity, and their incircles are congruent, $\triangle ABE \cong \triangle DFA$ .

Let $a = AF = BE$ , $b = FD = AB$ , and $c = AD = AE$ , and extend $DC$ and $AE$ to meet at $G$ :

Then $\triangle GFD \sim \triangle DFA$ by AA similarity, $GF = FD \cdot \cfrac{FD}{AF} = b \cdot \cfrac{b}{a} = \cfrac{b^2}{a}$ and $GD = FD \cdot \cfrac{AD}{FD} = b \cdot \cfrac{c}{a} = \cfrac{bc}{a}$ .

The inradius $r$ of right $\triangle GFD$ is $r = \cfrac{1}{2}(FD + GF - GD) = \cfrac{1}{2}\bigg(b + \cfrac{b^2}{a} - \cfrac{bc}{a}\bigg) = \cfrac{3}{4}$ .

By the Pythagorean Theorem, $a = \sqrt{c^2 - b^2}$ , and substituting this into $\cfrac{1}{2}\bigg(b + \cfrac{b^2}{a} - \cfrac{bc}{a}\bigg) = \cfrac{3}{4}$ and rearranging gives $(4b - 3)(24c - 16b^2 + 12b - 9) = 27$ .

Since rectangle $ABCD$ has integer sides, $b$ and $c$ must be positive integers, which means $4b - 3$ is a positive integer, and from the equation above, $4b - 3$ must also be a factor of $27$ , so either $4b - 3 = 1$ , $4b - 3 = 3$ , $4b - 3 = 9$ , or $4b - 3 = 27$ .

If $4b - 3 = 1$ , then $b = 1$ , but $c = \cfrac{5}{3}$ , a non-integer.

If $4b - 3 = 3$ , then $b = \cfrac{3}{2}$ , a non-integer.

If $4b - 3 = 27$ , then $b = \cfrac{15}{2}$ , a non-integer.

If $4b - 3 = 9$ , then $b = 3$ and $c = 5$ , both integers.

Therefore, $b = 3$ , $c = 5$ , and the area of rectangle $ABCD$ is $\boxed{15}$ .