2018 AMC 10B Problem 17

P16-20~6 minPrint

Triangles: Area & Pythagorean Quadratics Set up the equation/formula and compute; no special trick needed Exploit symmetry to reduce work or pair up objects

In rectangle $PQRS$ , $PQ=8$ and $QR=6$ . Points $A$ and $B$ lie on $\overline{PQ}$ , points $C$ and $D$ lie on $\overline{QR}$ , points $E$ and $F$ lie on $\overline{RS}$ , and points $G$ and $H$ lie on $\overline{SP}$ so that $AP=BQ<4$ and the convex octagon $ABCDEFGH$ is equilateral. The length of a side of this octagon can be expressed in the form $k+m\sqrt{n}$ , where $k$ , $m$ , and $n$ are integers and $n$ is not divisible by the square of any prime. What is $k+m+n$ ?

Show solution without answering

Key insight

The octagon is the rectangle with four matching right triangles cut off its corners. Write both legs using the side length, then use the Pythagorean theorem.

Solution

Call the octagon's side $s$ . The octagon is the rectangle with a right triangle cut off each corner, and each cut edge is a side of the octagon, so every corner triangle has long side $s$ .

1. Name the legs. Let $x = AP = BQ$ (given equal). Let $y = CQ$ .
2. Along the top, $PQ = 8$ . The side $AB$ is what is left of $PQ$ after the two pieces $x$ :
$s = 8 - 2x, \qquad\text{so}\qquad x = \frac{8 - s}{2}.$
3. The four corner triangles match. The triangles at $P$ and $Q$ both have one leg $x$ and long side $s$ , so their other legs match too: $HP = CQ = y$ . Going around the rectangle the same way, the triangles at $R$ and $S$ also match these, so every corner triangle has legs $x$ and $y$ .
4. Along the side, $QR = 6$ . The side $CD$ is what is left of $QR$ after the two pieces $y$ :
$s = 6 - 2y, \qquad\text{so}\qquad y = \frac{6 - s}{2}.$
5. Pythagorean theorem in a corner triangle.
$\left(\frac{8-s}{2}\right)^2 + \left(\frac{6-s}{2}\right)^2 = s^2.$
Multiply by $4$ to clear the fractions:
$(8-s)^2 + (6-s)^2 = 4s^2.$
Expand the left side: $64 - 16s + s^2 + 36 - 12s + s^2 = 100 - 28s + 2s^2$ . So
$100 - 28s + 2s^2 = 4s^2 \quad\Longrightarrow\quad s^2 + 14s - 50 = 0.$
6. Solve by completing the square. Rewrite as $s^2 + 14s = 50$ and add $49$ to both sides: $s^2 + 14s + 49 = 99$ , so $(s + 7)^2 = 99$ . The side is positive, so
$s = -7 + \sqrt{99} = -7 + 3\sqrt{11}.$
7. Check it makes sense. $s \approx -7 + 9.95 = 2.95$ , so $x \approx 2.5$ , which is less than $4$ as required.

So $k = -7$ , $m = 3$ , $n = 11$ , and $k + m + n = 7$ .

The answer is $\boxed{\textbf{(B)}\ 7}$ .

Why this works

An octagon with equal sides sitting in a rectangle is just the rectangle with four right-triangle corners removed. The rectangle's length and width each give a simple equation for one leg, and the Pythagorean theorem ties them together. The word "equilateral" (equal sides) rather than "regular" is the hint that the angles are not all equal.

Alternative approach

Solve $s^2 + 14s - 50 = 0$ with the quadratic formula instead:
$s = \frac{-14 + \sqrt{14^2 + 4 \cdot 50}}{2} = \frac{-14 + \sqrt{396}}{2} = \frac{-14 + 6\sqrt{11}}{2} = -7 + 3\sqrt{11}.$
A quick numeric check of the setup: with $s \approx 2.95$ , the legs are $x \approx 2.52$ and $y \approx 1.52$ , and $2.52^2 + 1.52^2 \approx 8.66 \approx 2.95^2$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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 $\boxed{\textbf{(B)}\ 7}$ .

Review in the AoPS books

Introduction to Geometry, Chapter 4: Perimeter and Area (§4.2)
Introduction to Algebra, Chapter 10: Quadratic Equations – Part 1 (§10.2, §10.3, §10.4)
Introduction to Algebra, Chapter 22: Special Manipulations (§22.3)
Introduction to Geometry, Chapter 6: Right Triangles (§6.1, §6.2, §6.3)