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

△ ABC has a right angle at C and hypotenuse AB = 12. A square ABXY is built outward on the hypotenuse, and a square ACWZ is built outward on leg AC. The four outer square-corners X, Y, Z, W all sit on one common circle. Find the perimeter AC + BC + AB of the triangle.

Pick an answer.

(A)

$12+9\sqrt{3}$

(B)

$18+6\sqrt{3}$

(C)

$12+12\sqrt{2}$

(D)

(E)

AMC 10 2015 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

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

Let b = AC and a = BC be the legs; with the right angle at C and AB = 12, Pythagoras gives a² + b² = 144.

a² + b² = AB² = 12² = 144

2STEP 2

Drop everything onto a coordinate grid

Put C = (0,0), A = (b,0), B = (0,a); building each square outward gives W=(0,-b), Z=(b,-b), X=(a,a+b), Y=(a+b,b).

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

Each perpendicular bisector matches one for triangle ABC, so the center is its circumcenter — the hypotenuse midpoint M = (b/2, a/2).

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

4STEP 4

Force the four corners onto one radius

Squared distances from M: to a hypotenuse corner MX² = 180, to a leg corner MW² = a²/4 + 5b²/4 + ab; one circle needs MW² = MX².

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

5STEP 5

Solve the two equations together

Substituting a²+b²=144 into a²+5b²+4ab=720 gives b²+ab=144=a²+b², so ab=a² and a = b = 6√(2) — an isosceles right triangle.

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

6STEP 6

Add up the three sides

With AC = BC = 6√(2) and AB = 12, the two legs give 12√(2), so 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