AMC 10 · 2010 · #19
Grade 8 geometry-2dA circle with center O has area 156π. Triangle ABC is equilateral, BC is a chord on the circle, OA=43, and point O is outside △ABC. What is the side length of △ABC?
Pick an answer.
AMC 10 2010 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 circle centered at $O$ has area $156\pi$. An equilateral triangle $ABC$ has side $\overline{BC}$ sitting as a chord of the circle, its third vertex $A$ off to one side, and the center $O$ lands outside the triangle. Given $OA = 4\sqrt{3}$, find the side length of $\triangle ABC$.
Givens: The circle has center $O$ and area $156\pi$; $\triangle ABC$ is equilateral; $\overline{BC}$ is a chord of the circle (so $B$ and $C$ lie on the circle); $OA = 4\sqrt{3}$; Point $O$ is outside $\triangle ABC$; Answer choices: (A) $2\sqrt{3}$, (B) $6$, (C) $4\sqrt{3}$, (D) $12$, (E) $18$
Unknowns: The side length of $\triangle ABC$
Understand
Restated: A circle centered at $O$ has area $156\pi$. An equilateral triangle $ABC$ has side $\overline{BC}$ sitting as a chord of the circle, its third vertex $A$ off to one side, and the center $O$ lands outside the triangle. Given $OA = 4\sqrt{3}$, find the side length of $\triangle ABC$.
Givens: The circle has center $O$ and area $156\pi$; $\triangle ABC$ is equilateral; $\overline{BC}$ is a chord of the circle (so $B$ and $C$ lie on the circle); $OA = 4\sqrt{3}$; Point $O$ is outside $\triangle ABC$; Answer choices: (A) $2\sqrt{3}$, (B) $6$, (C) $4\sqrt{3}$, (D) $12$, (E) $18$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #4 Introduce a Variable, #7 Identify Subproblems, #3 Eliminate Possibilities
The problem is a picture waiting to be drawn, so Tool #1 (Draw a Diagram) leads: once the figure is on paper, one clean fact appears — the altitude of the equilateral triangle and the center-to-chord perpendicular are the SAME line, so $A$, the midpoint $M$ of $BC$, and $O$ all sit on one straight line. Tool #4 (Introduce a Variable) names the side $s$ so lengths become expressions. Tool #7 (Identify Subproblems) splits the work into two right triangles — one for the radius-chord relation, one for the triangle's altitude. Tool #3 (Eliminate Possibilities) then uses the 'outside' clue to pick the correct side length from the two values the algebra allows.
Execute — Answer: B
7.G.A.2 Step 1 Draw it and find the straight line
- Sketch the circle, the chord $BC$, and vertex $A$.
- Let $M$ be the midpoint of $BC$.
- Because $\triangle ABC$ is equilateral, the segment $AM$ is its altitude and $AM \perp BC$.
- Because $BC$ is a chord, the line from center $O$ perpendicular to $BC$ also passes through $M$.
- Both facts point at the same perpendicular through $M$, so $A$, $M$, and $O$ lie on one straight line.
💡 Everything in the picture is symmetric across the perpendicular bisector of $BC$, so that one line holds the vertex, the chord's midpoint, and the center all at once.
7.G.B.4 Step 2 Turn the area into the radius
- The area of a circle is $\pi r^2$.
- Set it equal to the given area and solve for $r^2$; we only ever need $r^2$, not $r$ itself.
💡 The area formula is the only bridge from '$156\pi$' to a length, and $r^2$ is exactly what the Pythagorean step will ask for.
8.G.B.7 Step 3 Right triangle on the chord
- Let $s$ be the side length.
- Then $M$ is the midpoint of $BC$, so $BM = \tfrac{s}{2}$, and $B$ is on the circle, so $OB = r$.
- Triangle $OMB$ is right-angled at $M$.
- By the Pythagorean Theorem, find $OM$.
💡 Dropping the perpendicular to the chord builds a right triangle whose legs are the half-chord and the center's distance, tied together by $r$.
8.G.B.7 Step 4 Altitude of the equilateral triangle
- In equilateral $\triangle ABC$, the altitude $AM$ splits it into a right triangle with hypotenuse $s$ and one leg $\tfrac{s}{2}$.
- The Pythagorean Theorem gives the altitude.
💡 The same right-triangle trick that measures the chord also measures the triangle's height — half the base and the full side pin down the altitude.
8.G.B.7 Step 5 Use 'outside' to build one equation
- Since $A$, $M$, $O$ are collinear, $OA$ is just a piece of that line.
- If $O$ were the interior point between $M$ and $A$, it would sit inside the triangle — but $O$ is outside, so $O$ lies beyond $A$, giving $OM = OA + AM$.
- Substitute $OA = 4\sqrt{3}$ and $AM = \tfrac{\sqrt{3}}{2}s$, then square and set equal to $156 - \tfrac{s^2}{4}$ from the chord step.
💡 The word 'outside' is not decoration — it fixes $O$ on the far side of $A$, which is the only way to know how the pieces $OA$ and $AM$ add up.
6.EE.B.5 Step 6 Check which side length fits
- Clean up the equation: the $s^2$ terms combine to give $s^2 + 12s - 108 = 0$, which factors as $(s - 6)(s + 18) = 0$.
- A length cannot be negative, so $s = 6$.
- Test it against the answer choices: $s = 6$ makes $OM = 4\sqrt{3} + 3\sqrt{3} = 7\sqrt{3}$, and $OM^2 = 147 = 156 - 9$ matches the chord relation exactly.
- The trap value $s = 18$ is the case where $O$ falls inside the triangle, which the problem forbids.
- So the side length is $6$, choice $\textbf{(B)}$.
💡 Two candidates survive the algebra; the 'outside' condition is the referee that throws out $18$ and keeps $6$.
7.G.A.2 Sketch the circle, the chord $BC$, and vertex $A$. Let $M$ be the midpoint of $B 7.G.B.4 The area of a circle is $\pi r^2$. Set it equal to the given area and solve for 8.G.B.7 Let $s$ be the side length. Then $M$ is the midpoint of $BC$, so $BM = \tfrac{s} 8.G.B.7 In equilateral $\triangle ABC$, the altitude $AM$ splits it into a right triangl 8.G.B.7 Since $A$, $M$, $O$ are collinear, $OA$ is just a piece of that line. If $O$ wer 6.EE.B.5 Clean up the equation: the $s^2$ terms combine to give $s^2 + 12s - 108 = 0$, wh Review
Reasonableness: Plug $s = 6$ all the way back. Radius: $r^2 = 156$, so $r = 2\sqrt{39} \approx 12.49$. Half-chord $BM = 3$, so $OM = \sqrt{156 - 9} = \sqrt{147} = 7\sqrt{3} \approx 12.12$. Altitude $AM = \tfrac{\sqrt{3}}{2}(6) = 3\sqrt{3} \approx 5.20$. Since $A$, $M$, $O$ are in a line with $O$ past $A$, $OA = OM - AM = 7\sqrt{3} - 3\sqrt{3} = 4\sqrt{3}$, exactly the given value. And $OM \approx 12.12 > AM \approx 5.20$ confirms $O$ is beyond $A$, hence outside the triangle. Everything closes, and $6$ is choice (B).
Alternative: Skip the algebra and use Tool #3 (Eliminate Possibilities) directly on the choices. For each candidate $s$, compute $OM = \sqrt{156 - s^2/4}$ from the chord and $AM = \tfrac{\sqrt{3}}{2}s$ from the altitude, then check whether $OM - AM$ (the 'outside' case) equals $4\sqrt{3}$. Only $s = 6$ gives $7\sqrt{3} - 3\sqrt{3} = 4\sqrt{3}$; the others miss. This confirms (B) with arithmetic alone, no quadratic needed.
CCSS standards used (min grade 8)
7.G.A.2Draw geometric shapes with given conditions including triangles (Drawing the circle, chord, and equilateral triangle to reveal that vertex $A$, the chord midpoint $M$, and center $O$ all lie on one straight line.)7.G.B.4Know the formulas for area and circumference of a circle (Turning the area $156\pi$ into $r^2 = 156$ using $\text{area} = \pi r^2$.)8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Finding $OM^2 = 156 - \tfrac{s^2}{4}$ from the radius-chord right triangle, the altitude $AM = \tfrac{\sqrt{3}}{2}s$ from the equilateral triangle, and combining them into $s^2 + 12s - 108 = 0$.)6.EE.B.5Understand solving an equation or inequality as a process of finding values (Testing which side length satisfies the equation and the 'outside' condition, keeping $s = 6$ and rejecting $s = 18$.)
⭐ Draw the picture: the triangle's height and the line from the center to the chord are the same line, so the vertex, the midpoint, and the center line up — then two right triangles and the word 'outside' pin the side length to $6$.
⭐ Draw the picture: the triangle's height and the line from the center to the chord are the same line, so the vertex, the midpoint, and the center line up — then two right triangles and the word 'outside' pin the side length to $6$.
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