AMC 10 · 2024 · #14
Grade 8 geometry-2dProblem
One side of an equilateral triangle of height 24 lies on line ℓ. A circle of radius 12 is tangent to line l and is externally tangent to the triangle. The area of the region exterior to the triangle and the circle and bounded by the triangle, the circle, and line ℓ can be written as ab−cπ, where a, b, and c are positive integers and b is not divisible by the square of any prime. What is a+b+c?
Pick an answer.
AMC 10 2024 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: An equilateral triangle of height $24$ has one side on line $\ell$. A circle of radius $12$ sits on the same side of $\ell$ as the triangle, tangent to $\ell$ and externally tangent to one slanted side of the triangle. The region squeezed between the triangle, the circle, and the line near the vertex where they meet has area $a\sqrt{b} - c\pi$ with $b$ squarefree. Find $a + b + c$.
Givens: Equilateral triangle with height $24$, one side along line $\ell$; Circle of radius $r = 12$ on the same side of $\ell$ as the triangle; Circle is tangent to $\ell$ and externally tangent to a slanted side of the triangle; The target region is bounded by the triangle, the circle, and the line $\ell$; Area is written in the form $a\sqrt{b} - c\pi$ with $a, b, c$ positive integers and $b$ squarefree; Answer choices: (A) $72$, (B) $73$, (C) $74$, (D) $75$, (E) $76$
Unknowns: The sum $a + b + c$
Understand
Restated: An equilateral triangle of height $24$ has one side on line $\ell$. A circle of radius $12$ sits on the same side of $\ell$ as the triangle, tangent to $\ell$ and externally tangent to one slanted side of the triangle. The region squeezed between the triangle, the circle, and the line near the vertex where they meet has area $a\sqrt{b} - c\pi$ with $b$ squarefree. Find $a + b + c$.
Givens: Equilateral triangle with height $24$, one side along line $\ell$; Circle of radius $r = 12$ on the same side of $\ell$ as the triangle; Circle is tangent to $\ell$ and externally tangent to a slanted side of the triangle; The target region is bounded by the triangle, the circle, and the line $\ell$; Area is written in the form $a\sqrt{b} - c\pi$ with $a, b, c$ positive integers and $b$ squarefree; Answer choices: (A) $72$, (B) $73$, (C) $74$, (D) $75$, (E) $76$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems
The problem describes the picture in words, so the first move is Tool #1 (Draw a Diagram): place line $\ell$ horizontally, drop the triangle on top, and put the circle in the wedge outside the triangle at the vertex $C$ where one slanted side meets $\ell$. The picture immediately shows that the target region is the wedge between two tangent segments and an arc. That target naturally splits into two clean subproblems (Tool #7): the kite-shaped quadrilateral formed by the vertex, the two tangent points, and the circle's center, minus the circular sector cut out of that quadrilateral. Each subproblem is a one-formula calculation, and subtracting gives the requested $a\sqrt{b} - c\pi$.
Execute — Answer: D
7.G.B.6 Step 1 - Draw and label.
- Put the triangle's base on $\ell$ with vertex $C$ on the right.
- The circle sits outside the triangle, on the same side of $\ell$, hugging the slanted side $CA$ near $C$.
- Let $E$ be the tangent point on $\ell$, $D$ the tangent point on $CA$, and $O$ the circle's center.
- The target region is bounded by segment $CE$ (on $\ell$), segment $CD$ (on side $CA$), and arc $DE$ (on the circle).
- The quadrilateral $ODCE$ has the target region as the part outside the sector $ODE$.
💡 Naming the four key points turns a fuzzy region into a Grade 7 "polygon minus circular piece" decomposition.
7.G.B.5 Step 2 - Read off the wedge angle at $C$.
- The triangle's interior angle at $C$ is $60^\circ$, so the supplementary angle on the other side of side $CA$ — the wedge that contains the circle — is $180^\circ - 60^\circ = 120^\circ$.
- By symmetry, the center $O$ lies on the bisector of this $120^\circ$ wedge, splitting it into two $60^\circ$ halves.
- So $\angle OCE = 60^\circ$.
💡 Supplementary angles on a straight line and the bisector property of a tangent circle are Grade 7 angle facts.
8.G.B.7 Step 3 - Subproblem A: the kite $ODCE$.
- The radii $OE$ and $OD$ meet the tangent lines at right angles, so $\triangle OEC$ is right-angled at $E$ with $OE = 12$ and $\angle OCE = 60^\circ$.
- That makes $\triangle OEC$ a $30$-$60$-$90$ triangle.
- Using the standard $30$-$60$-$90$ side ratio $1 : \sqrt{3} : 2$, the leg opposite $60^\circ$ is $\sqrt{3}$ times the leg opposite $30^\circ$.
- Here $OE = 12$ sits opposite $60^\circ$ and $CE$ sits opposite $30^\circ$, so $CE = 12/\sqrt{3} = 4\sqrt{3}$.
- Two such congruent right triangles ($OEC$ and $ODC$) make up the kite.
💡 The Grade 8 Pythagorean machinery gives the $30$-$60$-$90$ side ratio $1 : \sqrt{3} : 2$, which converts one known leg into the other in a single step.
7.G.B.5 Step 4 - Subproblem B: the sector $ODE$.
- The four angles of quadrilateral $ODCE$ add to $360^\circ$.
- Two of them are the right angles at the tangent points ($\angle ODC = \angle OEC = 90^\circ$), one is the wedge $\angle DCE = 120^\circ$, and the remaining angle is the central angle $\angle DOE$.
💡 The angle sum in a quadrilateral is the dual of "all the way around a point" — a Grade 7 angle bookkeeping move.
7.G.B.4 Step 5 - Apply the sector-area formula.
- A sector of central angle $\theta^\circ$ in a circle of radius $r$ has area $\tfrac{\theta}{360}\pi r^2$.
- Here $\theta = 60$ and $r = 12$, so the sector takes up one sixth of the full disk.
💡 The sector formula is just the Grade 7 circle-area formula $A = \pi r^2$ scaled by the angle fraction $\theta/360$.
6.EE.A.2 Step 6 - Subtract and match the answer form.
- The target area is the kite minus the sector.
- Comparing with $a\sqrt{b} - c\pi$ gives $a = 48$, $b = 3$, $c = 24$.
- The number $b = 3$ is squarefree as required, and $a, b, c$ are positive integers.
- Sum them.
💡 Reading coefficients off an expression of the form $a\sqrt{b} - c\pi$ is a Grade 6 "identify parts of an expression" move, then add.
7.G.B.6 Draw and label. Put the triangle's base on $\ell$ with vertex $C$ on the right. 7.G.B.5 Read off the wedge angle at $C$. The triangle's interior angle at $C$ is $60^\ci 8.G.B.7 Subproblem A: the kite $ODCE$. The radii $OE$ and $OD$ meet the tangent lines at 7.G.B.5 Subproblem B: the sector $ODE$. The four angles of quadrilateral $ODCE$ add to $ 7.G.B.4 Apply the sector-area formula. A sector of central angle $\theta^\circ$ in a cir 6.EE.A.2 Subtract and match the answer form. The target area is the kite minus the sector Review
Reasonableness: Sanity check the size of the wedge: $48\sqrt{3} \approx 48(1.732) \approx 83.1$, and $24\pi \approx 75.4$, so the target region has area roughly $7.7$. That is a small sliver — exactly what you would expect for the thin pocket between a circle of radius $12$ and a $60^\circ$ vertex of a triangle, so the order of magnitude is right. Each tangent length $CE = CD = 4\sqrt{3} \approx 6.93$, comfortably less than the radius $12$, which matches the picture where the tangent points sit closer to $C$ than $O$ does. The triangle's height $24$ never entered the calculation, consistent with the local nature of the region.
Alternative: Tool #3 (Eliminate Possibilities) on the form $a\sqrt{b} - c\pi$: every step except the final $30$-$60$-$90$ ratio and the sector formula is forced. Once a student knows that the sector angle is $60^\circ$ (a third of a half-disk and a sixth of a full disk are the only "nice" central angles that fit), the $\pi$ coefficient must be $\tfrac{60}{360}\cdot 144 = 24$, forcing $c = 24$. The $\sqrt{}$ part must come from a $30$-$60$-$90$ triangle, so $b = 3$ is forced; then $a$ is read off from one short kite calculation. Adding gives $a + b + c = 75$, matching (D).
CCSS standards used (min grade 8)
7.G.B.4Know the formulas for the area and circumference of a circle and use them to solve problems (Computing the sector area as $\tfrac{60}{360} \cdot \pi (12)^2 = 24\pi$, i.e. one sixth of the full disk.)7.G.B.5Use facts about supplementary, complementary, vertical, and adjacent angles to solve multi-step problems (Turning the $60^\circ$ interior angle of the equilateral triangle into the $120^\circ$ exterior wedge at vertex $C$, then using the quadrilateral angle sum to extract the $60^\circ$ central angle of the sector.)7.G.B.6Solve problems involving area of two-dimensional objects composed of triangles, quadrilaterals, and polygons (Decomposing the target region as the kite $ODCE$ minus the circular sector $ODE$, and computing each piece separately.)8.G.B.7Apply the Pythagorean Theorem to determine unknown side lengths in right triangles (Using the $30$-$60$-$90$ side ratio $1 : \sqrt{3} : 2$ (a direct Pythagorean consequence) to convert the radius $OE = 12$ into the tangent length $CE = 4\sqrt{3}$.)6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Matching the computed area $48\sqrt{3} - 24\pi$ to the template $a\sqrt{b} - c\pi$ to read off $a = 48$, $b = 3$, $c = 24$, then summing.)
⭐ Draw the picture, spot the kite plus pie-slice at the vertex, and the AMC 10 problem reduces to one $30$-$60$-$90$ triangle and one $\tfrac{1}{6}$-of-a-circle sector — Grade 7-8 geometry the whole way.
⭐ Draw the picture, spot the kite plus pie-slice at the vertex, and the AMC 10 problem reduces to one $30$-$60$-$90$ triangle and one $\tfrac{1}{6}$-of-a-circle sector — Grade 7-8 geometry the whole way.
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