AMC 10 · 2022 · #15
Grade 8 geometry-2dProblem
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 caπ−b, where a,b, and c are positive integers such that a and c have no common prime factor. What is a+b+c?
Pick an answer.
AMC 10 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
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
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) $1997$
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
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 $CD = 20, DA = 15$).
💡 Grade 7 geometric construction: adding the diagonal turns one complicated shape into two familiar triangles.
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.
💡 Grade 8 Pythagorean reasoning: recognizing two famous triples sharing the same hypotenuse pins down every angle in the picture.
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 $r = AC / 2 = 25/2$.
💡 Grade 7 circle fact: a $90^\circ$ inscribed angle and the diameter are two ways of seeing the same thing (Thales).
7.G.B.4 Step 4 Circle area: $A_{\text{circle}} = \pi r^2 = \pi \cdot \left(\dfrac{25}{2}\right)^2 = \dfrac{625\pi}{4}$.
💡 Grade 7 circle area formula applied to $r = 25/2$.
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 $= 84 + 150 = 234$.
💡 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.
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 $(a\pi - b)/c$.
💡 Grade 6 fraction arithmetic: rewrite $234 = 936/4$ to align denominators, then subtract.
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 $a + b + c = 625 + 936 + 4 = 1565$.
💡 Grade 6 GCF / prime-factor check confirms the form is canonical, then straight addition gives the answer.
6.EE.B.5 Step 8 Match $1565$ to the choice list: option (D).
💡 Final close-out vs the multiple-choice list.
7.G.A.2 Draw the quadrilateral with sides $AB = 7, BC = 24, CD = 20, DA = 15$ and add di 8.G.B.7 Pattern-spot the famous Pythagorean triples. $7^2 + 24^2 = 49 + 576 = 625 = 25^2 7.G.B.4 Use the inscribed-angle / Thales fact: an inscribed angle that subtends a diamet 7.G.B.4 Circle area: $A_{\text{circle}} = \pi r^2 = \pi \cdot \left(\dfrac{25}{2}\right) 6.G.A.1 Quadrilateral area is the sum of the two right-triangle areas (each is $\tfrac{1 6.NS.B.3 Subtract to get the interior-circle / exterior-quadrilateral area, and put it ov 6.NS.B.4 Match form: $a = 625$, $b = 936$, $c = 4$. Verify the $\gcd$ condition: $625 = 5 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.5Understand 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.3Fluently 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.4Find 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.1Find 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.2Draw 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.4Know 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.7Apply 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 $a + b + c = 625 + 936 + 4 = 1565$.
⭐ 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 $a + b + c = 625 + 936 + 4 = 1565$.
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