2018 AMC 10B Problem 17

Let the octagon's side be $s$. Cutting the four corners of the rectangle leaves four right triangles, each with hypotenuse $s$. Let $x = AP = BQ$ and let $y = QC$ be the other leg of the corner triangle at $Q$.

Side $AB$ lies on $PQ$: $AB = 8 - 2x = s$, so $x = \frac{8 - s}{2}$.

The corner triangles at $P$ and $Q$ both have a leg $x$ and hypotenuse $s$, so their other legs are equal: $HP = CQ = y$. The same reasoning at the other corners shows all four corner triangles are congruent, so $CD = 6 - 2y = s$ and $y = \frac{6 - s}{2}$.

Pythagoras in a corner triangle:
$$\left(\frac{8-s}{2}\right)^2 + \left(\frac{6-s}{2}\right)^2 = s^2.$$
Multiply by $4$: $(8-s)^2 + (6-s)^2 = 4s^2$, i.e. $100 - 28s + 2s^2 = 4s^2$, so
$$s^2 + 14s - 50 = 0, \qquad s = \frac{-14 + \sqrt{196 + 200}}{2} = \frac{-14 + 6\sqrt{11}}{2} = -7 + 3\sqrt{11}.$$
(Check: $s \approx 2.95$, so $x \approx 2.5 < 4$ as required.)

Thus $k + m + n = -7 + 3 + 11 = 7$. The answer is $\boxed{\textbf{(B)}\ 7}$.

An equilateral octagon inscribed in a rectangle is a rectangle with four congruent right-triangle corners removed. Two linear relations (along the length and the width) express both legs in terms of $s$, and the Pythagorean theorem supplies the one quadratic needed. The word "equilateral," not "regular," is the warning that the angles are not all equal.

Complete the square instead of using the quadratic formula: $s^2 + 14s - 50 = 0$ becomes $(s + 7)^2 = 99$, so $s = -7 + \sqrt{99} = -7 + 3\sqrt{11}$. Numerically $s \approx 2.95$; then $AB = s$, $AP \approx 2.52$ and $QC \approx 1.52$, and $2.52^2 + 1.52^2 \approx 8.66 \approx s^2$, a quick check of the setup.

• Assuming the octagon is regular (all angles 135 degrees) and using 45-45-90 corner triangles, which contradicts the 8-by-6 rectangle.
• Leaving the answer as $-7 + \sqrt{99}$ and computing $-7 + 1 + 99 = 93$, forgetting that $n$ must be squarefree.