AMC 10 · 2022 · #15

Grade 8 geometry-2d

Archetype Compound Area by Decomposition / Difference

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

📏 Long solution 💡 3 insights

Problem

Quadrilateral $ABCD$ with side lengths $AB=7, BC=24, CD=20, DA=15$ is inscribed in a circle. The area interior to the circle but exterior to the quadrilateral can be written in the form $\frac{a\pi-b}{c},$ where $a,b,$ and $c$ are positive integers such that $a$ and $c$ have no common prime factor. What is $a+b+c?$

$\textbf{(A) } 260 \qquad \textbf{(B) } 855 \qquad \textbf{(C) } 1235 \qquad \textbf{(D) } 1565 \qquad \textbf{(E) } 1997$

Pick an answer.

(A)

260

(B)

855

(C)

1235

(D)

1565

(E)

1997

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Toolkit + CCSS Solution

Understand

Restated: 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 $\dfrac{a\pi - b}{c}$ where $a, b, c$ are positive integers and $\gcd(a, c)$ has no prime factor in common. Find $a + b + c$.

Givens: Cyclic quadrilateral $ABCD$ with sides $AB=7, BC=24, CD=20, DA=15$; Required area form: $\dfrac{a\pi - b}{c}$; Condition: $a$ and $c$ share no common prime factor; Answer choices: (A) $260$, (B) $855$, (C) $1235$, (D) $1565$, (E) $1997$

Unknowns: $a + b + c$

Understand

Plan

Primary tool: #1 Draw a Diagram

Secondary: #5 Look for a Pattern, #7 Identify Subproblems, #3 Eliminate Possibilities

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 $\angle B + \angle D = 90^\circ + 90^\circ = 180^\circ$, 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 $25/2$, (b) sum of two right-triangle areas, (c) subtraction to the difference form, (d) reading off $a, b, c$ and summing.

Execute — Answer: D

#1 Draw a Diagram 7.G.A.2 Step 1

Draw the quadrilateral with sides $AB = 7, BC = 24, CD = 20, DA = 15$ and add diagonal $AC$ — that splits $ABCD$ into $\triangle ABC$ (sides $AB = 7, BC = 24$) and $\triangle ACD$ (sides $CD = 20, DA = 15$).

$$\text{Split: } \triangle ABC \text{ (sides } 7, 24, AC\text{) and } \triangle ACD \text{ (sides } 15, 20, AC\text{)}$$

💡 Grade 7 geometric construction: adding the diagonal turns one complicated shape into two familiar triangles.

#5 Look for a Pattern 8.G.B.7 Step 2

Pattern-spot the famous Pythagorean triples.
$7^2 + 24^2 = 49 + 576 = 625 = 25^2$, so $7$-$24$-$25$ is a right-triple.
Also $15 = 3 \cdot 5, 20 = 4 \cdot 5$, so $15$-$20$-$25$ is the $3$-$4$-$5$ triple scaled by $5$, and $15^2 + 20^2 = 225 + 400 = 625 = 25^2$.
Both triangles share diagonal $AC$, and both calculations point to $AC = 25$.
So $\triangle ABC$ is right-angled at $B$ and $\triangle ACD$ is right-angled at $D$, with $AC = 25$ as the common hypotenuse.

$$7^2 + 24^2 = 25^2 \;\Rightarrow\; \angle B = 90^\circ; \quad 15^2 + 20^2 = 25^2 \;\Rightarrow\; \angle D = 90^\circ; \quad AC = 25$$

💡 Grade 8 Pythagorean reasoning: recognizing two famous triples sharing the same hypotenuse pins down every angle in the picture.

#7 Identify Subproblems 7.G.B.4 Step 3

Use the inscribed-angle / Thales fact: an inscribed angle that subtends a diameter is a right angle, and vice versa.
Both $\angle B$ and $\angle D$ are inscribed right angles subtending the chord $AC$, so $AC$ must be a diameter of the circle.
Radius $r = AC / 2 = 25/2$.

$$AC = \text{diameter} \;\Rightarrow\; r = \dfrac{25}{2}$$

💡 Grade 7 circle fact: a $90^\circ$ inscribed angle and the diameter are two ways of seeing the same thing (Thales).

#7 Identify Subproblems 7.G.B.4 Step 4

Circle area: $A_{\text{circle}} = \pi r^2 = \pi \cdot \left(\dfrac{25}{2}\right)^2 = \dfrac{625\pi}{4}$.

$$A_{\text{circle}} = \pi \cdot \dfrac{625}{4} = \dfrac{625\pi}{4}$$

💡 Grade 7 circle area formula applied to $r = 25/2$.

#7 Identify Subproblems 6.G.A.1 Step 5

Quadrilateral area is the sum of the two right-triangle areas (each is $\tfrac{1}{2} \cdot \text{leg}_1 \cdot \text{leg}_2$).
$\text{Area}(\triangle ABC) = \tfrac{1}{2} \cdot 7 \cdot 24 = 84$.
$\text{Area}(\triangle ACD) = \tfrac{1}{2} \cdot 15 \cdot 20 = 150$.
Total $= 84 + 150 = 234$.

$$A_{\text{quad}} = \tfrac{1}{2}(7)(24) + \tfrac{1}{2}(15)(20) = 84 + 150 = 234$$

💡 Grade 6 area-by-decomposition: triangle areas are $\tfrac{1}{2} \cdot \text{base} \cdot \text{height}$, and right triangles let the legs play those roles.

#7 Identify Subproblems 6.NS.B.3 Step 6

Subtract to get the interior-circle / exterior-quadrilateral area, and put it over a common denominator $4$ to read off $(a\pi - b)/c$.

$$\dfrac{625\pi}{4} - 234 = \dfrac{625\pi}{4} - \dfrac{936}{4} = \dfrac{625\pi - 936}{4}$$

💡 Grade 6 fraction arithmetic: rewrite $234 = 936/4$ to align denominators, then subtract.

#7 Identify Subproblems 6.NS.B.4 Step 7

Match form: $a = 625$, $b = 936$, $c = 4$.
Verify the $\gcd$ condition: $625 = 5^4$ has only the prime factor $5$; $4 = 2^2$ has only the prime factor $2$.
No shared prime — condition met.
Sum $a + b + c = 625 + 936 + 4 = 1565$.

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

💡 Grade 6 GCF / prime-factor check confirms the form is canonical, then straight addition gives the answer.

#3 Eliminate Possibilities 6.EE.B.5 Step 8

Match $1565$ to the choice list: option (D).

$$1565 \;\Rightarrow\; \textbf{(D)}$$

💡 Final close-out vs the multiple-choice list.

[1] #1 7.G.A.2 Draw the quadrilateral with sides $AB = 7, BC = 24, CD = 20, DA = 15$ and add di

[2] #5 8.G.B.7 Pattern-spot the famous Pythagorean triples. $7^2 + 24^2 = 49 + 576 = 625 = 25^2

[3] #7 7.G.B.4 Use the inscribed-angle / Thales fact: an inscribed angle that subtends a diamet

[4] #7 7.G.B.4 Circle area: $A_{\text{circle}} = \pi r^2 = \pi \cdot \left(\dfrac{25}{2}\right)

[5] #7 6.G.A.1 Quadrilateral area is the sum of the two right-triangle areas (each is $\tfrac{1

[6] #7 6.NS.B.3 Subtract to get the interior-circle / exterior-quadrilateral area, and put it ov

[7] #7 6.NS.B.4 Match form: $a = 625$, $b = 936$, $c = 4$. Verify the $\gcd$ condition: $625 = 5

[8] #3 6.EE.B.5 Match $1565$ to the choice list: option (D).

Review

Reasonableness: Magnitude check. Circle area $625\pi/4 \approx 490.9$; quadrilateral area $234$; difference $\approx 256.9$, 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 $1565$. ✓

Alternative: Tool #6 (Guess and Check) on the answer choices: only the form $(a\pi - b)/c$ with the gcd constraint produces specific $a, b, c$ triples. Choice (A) $260$ would need $a, b, c$ summing to $260$ — far too small to fit $625 + 936 + 4$. Choice (E) $1997$ would require $a$ much larger than $625$, but the only circle in sight has diameter $25$. Only (D) $1565$ is consistent with the Pythagorean-triple radius-$25/2$ picture.

CCSS standards used (min grade 8)

6.EE.B.5 Understand solving an equation or inequality as a process of finding values (Matching $a + b + c = 1565$ against the five answer choices to pick (D).)
6.NS.B.3 Fluently add, subtract, multiply, and divide multi-digit decimals (Subtracting $\dfrac{625\pi}{4} - \dfrac{936}{4}$ after rewriting $234$ over the common denominator $4$.)
6.NS.B.4 Find greatest common factor and least common multiple of two numbers (Verifying $\gcd(625, 4) = 1$ via prime factorizations $625 = 5^4$ and $4 = 2^2$.)
6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing (Composing the quadrilateral area as the sum of two right triangles using $\tfrac{1}{2} \cdot \text{leg}_1 \cdot \text{leg}_2$.)
7.G.A.2 Draw geometric shapes with given conditions including triangles (Drawing the cyclic quadrilateral $ABCD$ with the given side lengths and inserting diagonal $AC$ to split it into two triangles.)
7.G.B.4 Know the formulas for area and circumference of a circle (Computing the circle area $\pi r^2$ with $r = 25/2$, and using the inscribed-angle / Thales fact about diameters.)
8.G.B.7 Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Recognizing $7^2 + 24^2 = 25^2$ and $15^2 + 20^2 = 25^2$ to identify both triangles as right triangles with $AC = 25$.)

⭐ 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 $625\pi/4$ gives $(625\pi - 936)/4$, so $a + b + c = 625 + 936 + 4 = 1565$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 8)

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