AMC 10 · 2018 · #17

Grade 8 geometry-2d

Archetype Coordinate Plane Figure Computation └ Sub-archetype Figure Constraint Equations for Unknown Quantity

pythagorean-theoremquadratic-equations convert-to-algebraidentify-subproblems ↑ Prerequisites: pythagorean-theorem

📏 Long solution 💡 3 insights

Problem

A rectangle PQRS has sides PQ=8 and QR=6. Two points sit on each side, and joining them in order makes a convex eight-sided figure ABCDEFGH whose eight sides are all the same length. That common length can be written as k+m√(n) with n square-free. Find k+m+n.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

106

AMC 10 2018 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Convert to Algebra

Tool #13 (Convert to Algebra): the equilateral condition is the key fact, and it becomes a single equation once we name the side s and notice each corner of the rectangle is cut off by a small right triangle. Tool #1 (Draw a Diagram): sketching the rectangle with the four corner triangles shows which segments are the octagon's sides and which are the leftover pieces along each edge. Tool #4 (Introduce a Variable): calling the side length s lets us write every leftover edge piece in terms of s. Tool #7 (Identify Subproblems): the work splits into two clean parts — first solve for s, then read off k, m, n and add them.

1STEP 1

Sketch the rectangle and its corner triangles

Draw rectangle PQRS; at each corner a small right triangle is sliced off, and its slanted cut is one octagon side of common length s.

AB=BC=CD=…=s (all eight sides equal)

2STEP 2

Write the corner legs in terms of s

Splitting each edge symmetrically gives legs BQ= $\frac{8-s}{2}$ and QC= $\frac{6-s}{2}$ at corner Q, whose hypotenuse is the octagon side s.

BQ=(8-s)/2, QC=(6-s)/2, BC=s

3STEP 3

Apply the Pythagorean theorem at corner Q

Since triangle BQC is right-angled at Q, the Pythagorean theorem ties its legs $\frac{8-s}{2}$ and $\frac{6-s}{2}$ to the hypotenuse s in one equation.

((8-s)/2)²+((6-s)/2)²=s²

4STEP 4

Clear fractions and simplify to a quadratic

Clearing the denominators and expanding both squares collapses everything to the quadratic s²+14s-50=0.

(8-s)²+(6-s)²=4s² ⟹ s²+14s-50=0

5STEP 5

Solve for the side by completing the square

Completing the square gives (s+7)²=99, so s=-7+3√(11); the negative root is discarded and BQ= $\frac{8-s}{2}$ ≈2.52 obeys the AP=BQ condition.

(s+7)²=99 ⟹ s=-7+3√(11)

6STEP 6

Read off k, m, n and add

Reading s=-7+3√(11) as k+m√(n) gives k=-7, m=3, n=11 (square-free), so the sum is k+m+n=-7+3+11=7, choice (B).

k+m+n=(-7)+3+11=7 → (B)

Answer

The side s=-7+3√(11)≈ 2.95 is a sensible length: it is positive, smaller than both rectangle sides, and forces BQ=

\frac{8-s}{2}

≈ 2.52, which obeys the stated condition AP=BQ ＜ 4. The negative root was correctly thrown out. Plugging s≈ 2.95 back in, the corner legs are about 2.52 and 1.52, and 2.52²+1.52²≈ 6.36+2.31=8.67≈ 2.95², so the Pythagorean relation checks out. The sum k+m+n=7 is exactly choice (B); the trap answers come from sign slips on k=-7 (taking +7 would inflate the sum) or from mishandling √(99)=3√(11).

💡Key takeaway

Each rectangle corner is cut by a right triangle whose slanted edge is one octagon side, so the Pythagorean theorem turns 'all sides equal' into a quadratic that solves to s=-7+3√(11).

Sketch the rectangle and its corner triangles
Write the corner legs in terms of s
Apply the Pythagorean theorem at corner Q
Clear fractions and simplify to a quadratic
Solve for the side by completing the square
Read off k, m, n and add