AMC 10 · 2015 · #19

Grade 8 geometry-2d

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

pythagorean-theoremperpendicular-bisectorisosceles-triangle convert-to-algebra ↑ Prerequisites: pythagorean-theorem

📏 Long solution 💡 4 insights

Problem

Squares built outward on two sides put their four far corners on one circle. Find the triangle's perimeter.

Pick an answer.

(A)

$12+9\sqrt{3}$

(B)

$18+6\sqrt{3}$

(C)

$12+12\sqrt{2}$

(D)

(E)

How to solve

Strategy Draw a Diagram

The words describe shapes stacked on a right triangle, so the first move is tool #1 (Draw a Diagram): pin the figure onto a coordinate grid with the right angle at the origin. That turns every square-corner into a pair of coordinates. Tool #4 (Introduce a Variable) names the two unknown legs a and b so the one fact we are handed — the four corners are concyclic — can be written as an equation. Tool #7 (Identify Subproblems) splits the work: first locate the circle's center, then force the four corners to share one radius. That single condition pins the triangle's shape, and the perimeter falls out.

1STEP 1

Name the legs and apply Pythagoras

The right angle gives one equation at once.

a² + b² = AB² = 12² = 144

2STEP 2

Drop everything onto a coordinate grid

A grid places every corner.

C=(0,0), A=(b,0), B=(0,a); W=(0,-b), Z=(b,-b), X=(a,a+b), Y=(a+b,b)

3STEP 3

Find the circle's center

Symmetry puts the centre at a midpoint.

M = midpoint of AB = (b/2, a/2)

4STEP 4

Force the four corners onto one radius

Equal radii give a second equation.

MX² = 180, MW² = a²/4 + 5b²/4 + ab, MW² = MX²

5STEP 5

Solve the two equations together

Solving makes the two legs equal.

b² + ab = a² + b² → a = b; 2a² = 144 → a = b = 6√(2)

6STEP 6

Add up the three sides

The perimeter is 12+12√2, choice (C).

AC + BC + AB = 6√(2) + 6√(2) + 12 = 12 + 12√(2) → (C)

Answer

12+12√(2)

Check the radius is really shared once a = b = 6√(2), so a² = b² = 72 and ab = 72. Hypotenuse-square corner: MX² = 180. Leg-square corner: MW² = 72/4 + (5 · 72)/4 + 72 = 18 + 90 + 72 = 180. They match, so all four corners truly lie on one circle of radius √(180). The perimeter 12 + 12√(2) ≈ 28.97 is also sensible — a touch under 30, just as expected for legs near 8.49 with a hypotenuse of 12.

💡Key takeaway

Drop a geometry figure onto a coordinate grid, then turn 'these points share a circle' into 'these points share one distance to the center' — equal distances pin down the shape.

Name the legs and apply Pythagoras
Drop everything onto a coordinate grid
Find the circle's center
Force the four corners onto one radius
Solve the two equations together
Add up the three sides