AMC 10 · 2021 · #24

Grade 10 geometry-2d

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

inscribed-angletangent-circleschord-perpendicular-from-centercoordinate-geometry identify-subproblemsconvert-to-algebra ↑ Prerequisites: inscribed-angle

📏 Long solution 💡 3 insights

Problem

A semicircle sits on a diameter of length 14. A circle touches that diameter at one point and cuts the semicircle at two points. The chord joining those two points has length three root three, and the angle subtending it from the touch point is 60 degrees. Write the area of the triangle formed by those three points in the requested form and add the three whole numbers in it.

Pick an answer.

(A)

110

(B)

114

(C)

118

(D)

122

(E)

126

How to solve

Strategy Draw a Diagram

Tool #1 (Draw a Diagram) does the heavy lifting: once you draw both centers, the tangency, and the common chord QR, two facts jump out — the center of Ω is straight up from P, and both centers lie on the perpendicular bisector of QR, which makes three points collinear. Tool #4 (Introduce a Variable) names the one thing still unknown, the horizontal position of P. Tool #7 (Identify Subproblems) splits the work into four small jobs: size of Ω, position of P, equation of line QR, distance from P to that line. Tool #13 (Convert to Algebra) turns the picture into coordinates so the last two jobs are one-line computations. Tool #3 (Eliminate Possibilities) kills the second of the two candidate positions for P, because it would put P off the segment AB.

1STEP 1

Set up coordinates

Tangency puts the centre straight above the touch point.

Γ: x² + y² = 49, P = (p, 0), M = (p, ρ)

2STEP 2

Inscribed becomes central

The inscribed 60 becomes a central 120.

∠ QMR = 2∠ QPR = 120°

3STEP 3

Get the circle's radius

The chord and central angle give the radius.

sin 60° = QN/ρ = 3√(3)/2ρ; √(3)/2 = 3√(3)/2ρ → ρ = 3

4STEP 4

Distances from both centres

Measure both centres' distance to the chord.

QN² = (3√(3)/2)² = 27/4; MN = 3/2, ON = 13/2

5STEP 5

Locate the touch point

One of two candidates falls outside and is dropped.

√(p² + 9) ∈ {5, 8}; √(p²+9)=8 → |p| = √(55) ＞ 7 (rejected); √(p²+9)=5 → p = 4

6STEP 6

Find the chord's line and the height

Find the chord's line and the height to it.

8x + 6y = 65; d(P, QR) = (|32 - 65|)/(√(64 + 36)) = 33/10

7STEP 7

Compute the area and add

Computing and adding gives 122.

[PQR] = 1/2 · 3√(3) · 33/10 = 99√(3)/20; a+b+c = 99+3+20 = 122

Answer

122

Size check on the area. Every point of △ PQR lies on Ω, whose radius is 3, and among all triangles with a 60° angle at P inscribed in that circle the largest is the equilateral one, with area √(3)/4(3√(3))² = 27√(3)/4 ≈ 11.69. Our value 99√(3)/20 ≈ 8.57 is comfortably below that, and positive, so it is in range. Height check: the height 33/10 = 3.3 must be at most the largest distance from a point of Ω to line QR, which is MN + ρ = 3/2 + 3 = 4.5; indeed 3.3 ＜ 4.5. Position check: P = (4, 0) has |4| ＜ 7, so P really is on AB, and M = (4, 3) is 5 from O, so Ω pokes out past Γ's rim by 5 + 3 - 7 = 1, which is exactly why it cuts Γ at two points. Finally 99√(3)/20 is already in lowest terms, and 99 + 3 + 20 = 122 matches choice (D).

💡Key takeaway

This AMC 12 problem needs only Grade 10 circle geometry: tangency puts the small circle's center straight above P, the inscribed 60° angle forces its radius to be 3, the shared chord QR forces both centers onto one line so P = (4, 0), and subtracting the two circle equations hands you line QR — then area = 1/2 · 3√(3) · 33/10 = 99√(3)/20 and 99 + 3 + 20 = 122.

Set up coordinates and use tangency
Inscribed 60 becomes central 120
Get the radius of the circle
Distances from both centers to QR
Locate the tangency point P
Find line QR and the height from P
Compute the area and add