[Geometry]
Problem: Let $T=36$. Square $A B C D$ has area $T$. Points $M, N, O$, and $P$ lie on $\overline{A B}$, $\overline{B C}, \overline{C D}$, and $\overline{D A}$, respectively, so that quadrilateral $M N O P$ is a rectangle with $M P=2$. Compute $M N$.

Solution: Let $A M=a$ and $A P=b$, and let $s=\sqrt{T}$ be the side length of square $A B C D$. Then $M B=s-a$ and $D P=s-b$. Using the right angles of $M N O P$ and complementary acute angles in triangles $A M P, B N M$, $C O N$, and $D P O$, note that

$$
\angle A M P \cong \angle B N M \cong \angle C O N \cong D P O
$$

Also note that $\mathrm{m} \angle B M N=180^{\circ}-\left(90^{\circ}+\mathrm{m} \angle A M P\right)$, so it also follows that

$$
\angle B M N \cong \angle C N O \cong \angle D O P \cong A P M
$$

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Thus, by side-angle-side congruence, it follows that $\triangle A M P \cong \triangle C O N$ and $\triangle B N M \cong \triangle D P O$. Moreover, by side-angle-side similarity, it follows that $\triangle A M P \sim \triangle B N M \sim \triangle C O N \sim \triangle D P O$. Thus $B N=s-b, N C=b$, $C O=a$, and $O D=s-a$. The similarity relation implies $\frac{A M}{B N}=\frac{A P}{B M}$, so $\frac{a}{s-b}=\frac{b}{s-a}$. Cross-multiplying, rearranging, and simplifying yields $s(a-b)=(a+b)(a-b)$. Thus either $a=b$ or $s=a+b$. In the case where $a=b, A M=A P=\frac{2}{\sqrt{2}}=\sqrt{2}$, so $M N=(s-\sqrt{2}) \sqrt{2}=s \sqrt{2}-2$. With $T=36, s=6$, and the answer is thus $6 \sqrt{\mathbf{2}}-\mathbf{2}$. For completeness, it remains to verify that for this particular value of $s$, the case where $s=a+b$ is impossible. Applying the Pythagorean Theorem in $\triangle M A P$ yields $a^{2}+b^{2}=4$. Now if $s=6=a+b$, then by squaring, it would follow that $a^{2}+b^{2}+2 a b=36 \Longrightarrow 4+2 a b=36 \Longrightarrow a b=16$. But the equation $a+b=a+\frac{16}{a}=6$ has no real solutions, thus $a+b \neq 6$. (Alternatively, note that by the Arithmetic Mean-Geometric Mean Inequality, $a+\frac{16}{a} \geq 2 \sqrt{a \cdot \frac{16}{a}}=8>6$.)

Final answer: $6 \sqrt{2}-2$