Riemann surface
matlab

Consider a semi-circle with diameter $A B$. Let points $C$ and $D$ be on diameter $A B$ such that $C D$ forms the base of a square inscribed in the semicircle. Given that $C D=2$, compute the length of $A B$.

Solution: Note that the center of the semi-circle lies on the center of one of the sides of the square. If we draw a line from the center to an opposite corner of the square, we form a right triangle whose side lengths are 1 and 2 and whose hypotenuse is the radius of the semicircle. We can therefore use the Pythagorean Theorem to compute $r=\sqrt{1^2+2^2}=\sqrt{5}$. The radius is half the length of $A B$ so therefore $A B=2 \sqrt{5}$.

Let $A B C D$ be a trapezoid with $A B$ parallel to $C D$ and perpendicular to $B C$. Let $M$ be a point on $B C$ such that $\angle A M B=\angle D M C$. If $A B=3, B C=24$, and $C D=4$, what is the value of $A M+M D$ ?

Solution: Let $A^{\prime}$ be the reflection of $A$ by $B C$. We have $\angle A^{\prime} M B=\angle A M B=\angle D M C$. Hence, $A^{\prime}, M$, and $D$ are collinear. Let $C^{\prime}$ be the intersection of the line parallel to $B C$ passing through $A^{\prime}$ and the extension of $D C$. We have $\angle A^{\prime} C^{\prime} D=90^{\circ}, A^{\prime} C^{\prime}=B C=24$, and $C^{\prime} D=C^{\prime} C+C D=A^{\prime} B+C D=A B+C D=3+4=7$. Therefore, $A M+M D=$ $A^{\prime} M+M D=A^{\prime} D=\sqrt{A^{\prime} C^{\prime 2}+C^{\prime} D^2}=\sqrt{24^2+7^2}=25$ by Pythagorean Theorem.

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