AMC 10 · 2010 · #19
Grade 8 geometry-2dPick an answer.
AMC 10 2010 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Draw it and find the straight line
Sketch the circle, chord BC, and vertex A. Altitude AM and the perpendicular from O both hit M, the midpoint of BC, so A, M, O line up.
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.A.2Draw A DiagramTurn the area into the radius
A circle's area is π r², so matching it to the given area gives r² = 156 — r² is all the next step needs.
The area formula is the only bridge from '156π' to a length, and r² is exactly what the Pythagorean step will ask for.
7.G.B.4Identify SubproblemsRight triangle on the chord
Let s be the side. Then BM = s/2 and OB = r, and triangle OMB is right-angled at M, so Pythagoras gives OM² = 156 - s²/4.
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.
Dropping the perpendicular to the chord builds a right triangle whose legs are known lengths.
▸ Why?
The centre is equally far from both ends of the chord, so that line halves it at a right angle.
▸ Why?
That right angle ties the half-chord, the centre distance, and the radius into one equation.
Altitude of the equilateral triangle
The altitude AM cuts equilateral △ ABC into a right triangle with hypotenuse s and leg s/2, so AM = √(3)/2 s.
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.7Introduce A VariableUse 'outside' to build one equation
O outside means O sits beyond A on that line, so OM = OA + AM; substitute the two lengths, square, and match 156 - s²/4.
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.
8.G.B.7Introduce A VariableCheck which side length fits
The equation tidies to s² + 12s - 108 = 0, so (s - 6)(s + 18) = 0; a length is positive, leaving s = 6, choice (B).
Two candidates survive the algebra; the 'outside' condition is the referee that throws out 18 and keeps 6.
6.EE.B.5Eliminate PossibilitiesDraw 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 it and find the straight line
- Turn the area into the radius
- Right triangle on the chord
- Altitude of the equilateral triangle
- Use 'outside' to build one equation
- Check which side length fits