AMC 10 · 2019 · #23

Grade 8 geometry-2d

Archetype Coordinate Plane Figure Computation └ Sub-archetype Pythagorean / Distance Formula for a Length

pythagorean-theoremcoordinate-geometrychord-perpendicular-from-centerperpendicular-bisectorarea-circles identify-subproblems ↑ Prerequisites: pythagorean-theoremcoordinate-geometryarea-circles

📏 Long solution 💡 4 insights

Problem

Points A = (6, 13) and B = (12, 11) lie on a circle ω. The tangent lines to ω at A and B meet at a single point P on the x-axis. Find the area of ω.

Pick an answer.

(A)

$\frac{83\pi}{8}$

(B)

$\frac{21\pi}{2}$

(C)

$\frac{85\pi}{8}$

(D)

$\frac{43\pi}{4}$

(E)

$\frac{87\pi}{8}$

AMC 10 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

Tool #1 (Diagram): the picture (circle, two tangent points, external point on the x-axis) clarifies that |PA| = |PB| and that △ OAP is right-angled at A. Tool #7 (Subproblems): split into (a) find P on the x-axis from |PA| = |PB|, (b) compute |PA|, (c) use right-triangle/Ptolemy to relate r, |PA|, |PO|, (d) solve for r². Tool #13 (Algebra): each subproblem reduces to a short algebraic step (linear equation for t, distance squared, Pythagorean relation, Ptolemy).

1STEP 1

Draw A, B, circle ω, the two tangents meeting at P = (t, 0) on the x-axis, and center O; mark right angles at A and B.

P = (t, 0), OA ⊥ PA, OB ⊥ PB

2STEP 2

Equal tangents give (t - 6)² + 169 = (t - 12)² + 121, which simplifies to 12t = 60, so t = 5.

(t-6)² + 169 = (t-12)² + 121 → 12t = 60 → t = 5

3STEP 3

Then |PA|² = (6 - 5)² + 13² = 170, and |PB|² = 49 + 121 = 170 confirms it.

|PA|² = |PB|² = 170

4STEP 4

Chord |AB| = 2√(10) (from 36 + 4 = 40); with midpoint M = (9, 12), |PM|² = 160 gives |PM| = 4√(10).

|AB| = 2√(10), |PM| = 4√(10)

5STEP 5

PAOB is a cyclic kite (right angles at A, B), so Ptolemy gives 2√(170)·r = PO·2√(10), hence PO = r√(17).

2 √(170) · r = PO · 2√(10) → PO = r√(17)

6STEP 6

Pythagoras on right triangle OAP: (r√(17))² = r² + 170, so 16r² = 170, hence r² = $\frac{85}{8}$ .

17 r² = r² + 170 → r² =

\frac{85}{8}

7STEP 7

Area = π r² = $\frac{85\pi}{8}$ , which is choice (C).

Area = π r² =

\frac{85\pi}{8}

Answer

\frac{85\pi}{8}

Verify by independent coordinate computation. The center O lies on the perpendicular bisector of AB, which has slope -

\frac{1}{3}

, so the bisector has slope 3 and passes through M = (9, 12): y = 3x - 15. The center also lies on line PM from P = (5, 0) through M = (9, 12): that line is y = 3(x - 5) = 3x - 15 — same line ✓ (as it should be, since the locus of points equidistant from the two tangents through A, B is exactly the line through P and M). So O is on y = 3x - 15, and additionally on the line through A perpendicular to PA. PA direction is (1, 13), so the line through A = (6, 13) perpendicular to PA has direction (-13, 1): parametrically (6 - 13s, 13 + s). Plug into y = 3x - 15: 13 + s = 3(6 - 13s) - 15 = 3 - 39s, so 40s = -10, s = -

\frac{1}{4}

. Then O = (6 +

\frac{13}{4}

, 13 -

\frac{1}{4}

) = (

\frac{37}{4}

\frac{51}{4}

). Now r² = (

\frac{37}{4}

- 6)² + (

\frac{51}{4}

- 13)² = (

\frac{13}{4}

)² + (-

\frac{1}{4}

)² =

\frac{169 + 1}{16}

\frac{170}{16}

\frac{85}{8}

✓. Same value. Area =

\frac{85\pi}{8}

💡Key takeaway

This AMC 10 problem only needs Grade 8 Pythagoras and the circle's basic 'radius ⟂ tangent' fact — solve the linear equation |PA| = |PB| to find P = (5, 0), get |PA|² = 170, then 16 r² = 170 from a right-triangle/Ptolemy relation, so the area is $\frac{85\pi}{8}$ .

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