Quadrilateral Inscribed In Semicircle A

Label the vertices $A$ , $B$ , $C$ , and $D$ so that $AD = 105$ and $AB < BC < CD < AD$ , and draw diagonals $AC$ and $BD$ . Since $AB$ , $BC$ , and $CD$ are of integer lengths in an arithmetic progression, let $AB = t - k$ , $BC = t$ , and $CD = t + k$ , where $t$ and $k$ are positive integers. Also, since $BC < AD$ , that means $t < 105$ .

By Thales's Theorem, $\triangle ABD$ is a right triangle, and by Pythagorean's Theorem $BD = \sqrt{105^2 - (t - k)^2}$ . Similarly, $AC = \sqrt{105^2 - (t + k)^2}$ .

By Ptolemy's Theorem, $AC \cdot BD = AB \cdot CD + BC \cdot AD$ , which after substitution becomes $\sqrt{105^2 - (t - k)^2} \cdot \sqrt{105^2 - (t + k)^2} = (t - k)(t + k) + 105t$ and simplifies to $2k^2 = \frac{2t^3 + 315t^2 - 1157625}{t - 105} = \frac{(t + 105)^2(2t - 105)}{t - 105} = 2t^2 + 525t + 55125 + \frac{4630500}{t - 105}$ .

Since $AB = t - k > 0$ , that means $t > k$ and $2t^2 > 2k^2$ . Therefore, $2t^2 > 2t^2 + 525t + 55125 + \frac{4630500}{t - 105}$ , and since $t < 105$ , this solves to $t > \sqrt{2205} \geq 47$ .

Since $2k^2 > 0$ , $\frac{(t + 105)^2(2t - 105)}{t - 105} > 0$ , and since $t < 105$ , this means that $t < \frac{105}{2} \leq 52$ .

Since $2k^2$ is an integer and $2k^2 = 2t^2 + 525t + 55125 + \frac{4630500}{t - 105}$ , $t - 105$ must divide evenly into $4630500 = 2^23^35^37^3$ . Out of the integers in the range $47 \leq t \leq 52$ , only $t = 51$ gives a $t - 105$ value that divides evenly into $4630500$ .

Therefore, $t = 51$ , $k = \sqrt{\frac{(t + 105)^2(2t - 105)}{2(t - 105)}} = \sqrt{\frac{(51 + 105)^2(2\cdot 51 - 105)}{2(51 - 105)}} = 26$ , and $AB = t - k = 51 - 26 = \boxed{25}$ .

Since there is no other solution provided, I submit my numerical solution.

Let the three shorter side lengths of increasing length be $a$ , $b$ , and $c$ and the angles they extend at the center of the semicircle be $\alpha$ , $\beta$ , and $\gamma$ respectively. Then we have $a = 105 \sin \frac \alpha 2$ , $b = 105 \sin \frac \beta 2$ , and $c = 105 \sin \frac \gamma 2$ . Since $\frac \alpha 2 + \frac \beta 2 + \frac \gamma 2 = 90^\circ$ , we have:

$b = 105 \sin \frac \beta 2 = 105 \sin \left(90^\circ - \frac \alpha 2 - \frac \gamma 2\right) = 105 \cos \left(\frac \alpha 2 + \frac \gamma 2\right) = 105 \cos \left(\sin^{-1}\frac a{105} + \sin^{-1} \frac c{105} \right)$

We can find $b$ from $a$ and $c$ . We just have to cast a large enough integral net of $a$ and $c$ to get the integral $b$ . I used the following Microsoft spreadsheet for the net.

From the spreadsheet there are three integral solutions

$\begin{cases} a = 21 & b = 57 & c = 75 & \implies a + c \ne 2b & \red{\text{rejected}} \\ a = 25 & b = 51 & c = 77 & \implies a + c = 2b & \blue{\text{accepted}} \\ a = 30 & b = 55 & c = 70 & \implies a + c \ne 2b & \red{\text{rejected}} \end{cases}$

Therefore the length of the shortest side is $\boxed{25}$ .

The sides are $(25, 51, 77, 105)$