AMC 10 · 2022 · #15

Grade 8 geometry-2d

Archetype Compound Area by Decomposition / Difference └ Sub-archetype Circle-Polygon Hybrid

pythagorean-theoreminteger-pythagorean-triplesarea-trianglesarea-circles identify-subproblemsarea-differencepattern-recognition ↑ Prerequisites: pythagorean-theorem

📏 Long solution 💡 3 insights

Problem

A quadrilateral ABCD with side lengths AB = 7, BC = 24, CD = 20, DA = 15 sits inscribed in a circle. The region inside the circle but outside the quadrilateral has area equal to

\frac{a\pi - b}{c}

where a, b, c are positive integers and gcd(a, c) has no prime factor in common. Find a + b + c.

Pick an answer.

(A)

260

(B)

855

(C)

1235

(D)

1565

(E)

1997

AMC 10 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

Tool #1 (Draw): sketch the cyclic quadrilateral with sides 7, 24, 20, 15 in order and add diagonal AC — the picture immediately splits the shape into two triangles. Tool #5 (Pattern) is the key insight: 7-24-?? and 15-20-?? both scream familiar Pythagorean triples (7-24-25 and 3-4-5 scaled to 15-20-25). Both triangles share the diagonal of length exactly 25, which forces right angles at B and D. Since ∠ B + ∠ D = 90° + 90° = 180°, the quadrilateral is cyclic (consistent), and the right angles inscribed in a semicircle make AC a diameter. Tool #7 (Subproblems) then breaks the area calculation into (a) circle area from radius $\frac{25}{2}$ , (b) sum of two right-triangle areas, (c) subtraction to the difference form, (d) reading off a, b, c and summing.

1STEP 1

Add diagonal AC to split ABCD into two triangles: △ABC (sides 7, 24) and △ACD (sides 15, 20).

Split: △ ABC (sides 7, 24, AC) and △ ACD (sides 15, 20, AC)

2STEP 2

Both are famous Pythagorean triples: 7-24-25 and 15-20-25, so the shared diagonal is AC = 25, right-angled at B and D.

7² + 24² = 25² → ∠ B = 90°; 15² + 20² = 25² → ∠ D = 90°; AC = 25

3STEP 3

Right angles at B and D both subtend chord AC, so by Thales AC is a diameter and the radius is r = $\frac{25}{2}$ .

AC = diameter → r =

\frac{25}{2}

4STEP 4

Circle area = π r² = π · ( $\frac{25}{2}$ )² = $\frac{625\pi}{4}$ .

A_circle = π ·

\frac{625}{4}

\frac{625\pi}{4}

5STEP 5

Sum the two right-triangle areas: $\frac{1}{2}$ · 7 · 24 = 84 and $\frac{1}{2}$ · 15 · 20 = 150, so the quadrilateral area is 234.

A_quad =

\frac{1}{2}

(7)(24) +

\frac{1}{2}

(15)(20) = 84 + 150 = 234

6STEP 6

Subtract over denominator 4: $\frac{625\pi}{4}$ - 234 = $\frac{625\pi}{4}$ - $\frac{936}{4}$ = $\frac{625\pi - 936}{4}$ .

\frac{625\pi}{4}

- 234 =

\frac{625\pi}{4}

\frac{936}{4}

\frac{625\pi - 936}{4}

7STEP 7

Read off a = 625, b = 936, c = 4 (5⁴ and 2² share no prime factor), so a + b + c = 1565.

a + b + c = 625 + 936 + 4 = 1565

8STEP 8

1565 matches choice (D).

1565 → (D)

Answer

1565

Magnitude check. Circle area

\frac{625\pi}{4}

≈ 490.9; quadrilateral area 234; difference ≈ 256.9, which is the leftover area (a positive number, as expected for "inside circle minus inside quad"). The diameter 25 matches the picture too — the longest side is 24, just under the diameter, and the shortest is 7, leaving room for the chord; both fit inside a circle of diameter 25 comfortably. Final cross-check on 1565: 625 + 936 = 1561, plus 4 gives 1565. ✓

💡Key takeaway

This AMC 10 problem only needs Grade 8 Pythagorean-triple recognition you already know — the sides 7, 24 and 15, 20 form the famous triples 7-24-25 and 15-20-25. Both triangles share the same hypotenuse AC = 25, which must be the circle's diameter. Subtracting the quadrilateral area 234 from the circle area $\frac{625\pi}{4}$ gives $\frac{625\pi - 936}{4}$ , so a + b + c = 625 + 936 + 4 = 1565.

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