A geometry problem by Miles Koumouris

Let the diameter of the semicircle be $D$ and $\angle BCD = \theta$ . For a cyclic quadrilateral $ABCD$ , $\angle BAD = 180^\circ - \theta$ .

Using cosine rule on $\triangle BCD$ :

$\begin{aligned} {\color{#3D99F6} BD^2} & = BC^2 + CD^2 - 2 BC \cdot CD \cos \theta & \small \color{#3D99F6} \text{By Pythagorean theorem: } BD^2 = AD^2 - AB^2 \\ D^2 - 7^2 & = 58^2 + 92^2 - 2(58)(92) {\color{#3D99F6} \cos \theta} & \small \color{#3D99F6} \text{As } \cos (180^\circ - \theta) = \frac 7D, \ \cos \theta = - \frac 7D \\ D^2 - 49 & = 3364 + 8464 - 10672 \color{#3D99F6} \left(-\frac 7D \right) \\ D^2 & = 11877 + \frac {74707}D \end{aligned}$

$\begin{aligned} \implies \qquad D^3 -11877D -74704 & = 0 \\ (D-112)(D^2+112D+667) & = 0 \\ \implies D & = \boxed{112} & \small \color{#3D99F6} D^2 + 112D + 667 \text{ gives negative roots} \end{aligned}$

- Let the diameter of the semicircle be of length x. Since the quadrilateral is cyclic (It can be inscribed in a circle), by Ptolemy's Theorem, $58x+92 \cdot 7 = AC \cdot BD.$ Notice that the diagonals $AC$ and $BD$ form right triangles with the diameter, therefore, we can rewrite $AC$ as $\sqrt{x^2-92^2}$ and $BD$ as $\sqrt{x^2-7^2}.$ Replacing these into the equation gets us $58x+92 \cdot 7 = \sqrt{(x^2-92^2) \cdot (x^2-7^2)}.$ Squaring both sides gets us $3364x^2 + 74704x + 414736 = (x^2-92^2)(x^2-7^2)$ . From here, we solve for x. $\begin{aligned} 3364x^2+74704x+414736 = x^4-8513x^2+414736\\ x^4-11877x^2-74704x = 0 \\ x^3-11877x-74704 = 0 \\ (x-112)(x^2+112x+667) = 0 \\ \implies x = \boxed{112} \end{aligned}$

Let the radius of the semicircle be $r$ . From the diagram, it is evident that

$\textrm{sin}(\alpha )=\dfrac{7}{2r}$

$\textrm{sin}(\beta )=\dfrac{29}{r}$

$\textrm{sin}(\gamma )=\dfrac{46}{r}$ .

Since it is known that $2\alpha +2\beta +2\gamma =180^{\circ }\Longleftrightarrow \gamma =90^{\circ }-(\alpha +\beta )$ , we can say

$\textrm{sin}(\gamma )=\textrm{cos}(\alpha +\beta )$

$\textrm{cos}(\alpha +\beta )=\textrm{cos}(\alpha )\textrm{cos}(\beta )-\textrm{sin}(\alpha )\textrm{sin}(\beta )$ (by angle addition identities).

Letting $\textrm{sin}(\alpha )=\dfrac{7}{2r}=x$ , it is evident from trigonometric identites that

$\textrm{cos}(\alpha +\beta )=\sqrt{(1-x^{2})(1-(\dfrac{58x}{7})^{2})}-\dfrac{58}{7}x^{2}=\dfrac{92}{7}x$

$(7^{2}+58^{2}+92^{2})x^{2}+(2\times 58\times 92)x^{3}-7^{2}=0$ .

Substituting $x$ back as $\dfrac{7}{2r}$ gives

$4r^{3}-(7^{2}+58^{2}+92^{2})r-(7\times 58\times 92)=0$

$(r-56)(4r^{2}+224r+667)=0$ .

So $r=56$ , and thus the diameter of the semicircle is $\boxed{112}$ .

use ptolemy's equality then $\sqrt(D^2-92^2)$ $\sqrt(D^2-7^2)$ = $58D-92*7$ = $58D+644$ then square both side then simplifying we get $D^3-11877D-74704=0$ , then solve it by factorization $D=112$ :)

Let $p$ and $q$ be the shorter and longer diagonals of the quadrilateral and $d$ be the diameter of the semicircle. By Pythagorean and Ptolemy theorems,

$p^2+92^2 = d^2 = 7^2+q^2 \text{and } pq = 58d+7\cdot 92,$ $(58d+7\cdot 92)^2 = (pq)^2 = (d^2-92^2)(d^2-7^2),$ $d^4-(92^2+7^2+58^2)d^2-(2\cdot 58\cdot 7\cdot 92)d = 0,$ $(d-112)(d^2+112d+667) = 0,$ $d = 112 \text{ or } (d+56)^2 = 2469 < 50^2,$ But d > 0, so $d = 112.$