AMC 10 · 2020 · #8

Grade 8 geometry-2d

Archetype Casework by Position / Vertex Type └ Sub-archetype Vertex or Element Role

area-trianglesspatial-visualizationperpendicular-bisector caseworkidentify-subproblemssymmetry-argument ↑ Prerequisites: area-trianglespythagorean-theorem

📏 Long solution 💡 3 insights

Problem

P and Q are fixed points in the plane with PQ = 8. Count the locations of R that make △ PQR a right triangle of area 12.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2020 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

Tool #1 puts P and Q on a horizontal line so we can see the perpendicular conditions. Tool #7 splits the count into three cases by where the right angle sits (P, Q, or R). Tool #2 enumerates the positions in each case without missing reflections. Tool #3 matches the total against the choices.

1STEP 1

Put P, Q on a line 8 apart; area = $\frac{1}{2}$ · 8 · h = 4h = 12, so R must sit at height 3 from line PQ (one of two parallel lines).

\frac{1}{2}

· 8 · h = 12 → h = 3

2STEP 2

Case A — right angle at P: PR ⊥ PQ, and $\frac{1}{2}$ · 8 · PR = 12 gives PR = 3 → 2 spots, above and below P.

PR ⊥ PQ, PR = 3 → 2 positions

3STEP 3

Case B — right angle at Q: mirror of Case A, QR ⊥ PQ with QR = 3 → 2 more spots, above and below Q.

QR ⊥ PQ, QR = 3 → 2 positions

4STEP 4

Case C — right angle at R: PQ is the hypotenuse, so R lies on the circle with diameter PQ (center M, radius 4), still at height 3.

R on circle, center M, radius 4, height |y_R| = 3

5STEP 5

With M = (0,0), the circle is x² + y² = 16; y = ±3 gives x² = 7, x = ±√(7) — two points per line → 4 in Case C.

x² = 16 - 9 = 7 → x = ±√(7); 2 above + 2 below = 4

6STEP 6

All points are distinct (x = -4, 4, ±√(7)), so the cases just add: 2 + 2 + 4 = 8.

2 + 2 + 4 = 8

7STEP 7

8 matches choice (D).

8 → (D)

Answer

Symmetry check: the configuration is symmetric across line PQ (so the total must be even) and across the perpendicular bisector of PQ (the line x = 0). The eight points (-4, ± 3), (4, ± 3), (±√(7), ± 3) are closed under both reflections. ✓ Each point gives a triangle: Case A — legs 8, 3, area =

\frac{1}{2}

· 8 · 3 = 12 ✓; Case C — legs computed from Pythagorean theorem, PR² + QR² = 64 with PR · QR = 24 (since area = 12), consistent. ✓

💡Key takeaway

This AMC 10 problem only needs Grade 8 right-triangle thinking you already know — the area 12 forces R to be 3 units away from line PQ. Then the right angle can sit at P (2 spots), at Q (2 spots), or at R on the circle with PQ as diameter (4 spots). Total 8.

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