AMC 10 · 2008 · #9
Grade 8 geometry-2dPick an answer.
The problem is pure circle geometry, so tool #1 (Draw a Diagram) leads: sketching the center, the chord, and the arc midpoint exposes a line of symmetry that lines up the center O, the chord's midpoint M, and C. Once that line is drawn, the distance AC breaks into small right-triangle pieces, so tool #7 (Identify Subproblems) handles them in order — first the distance from the center to the chord, then the short gap MC, then AC itself. Tool #3 (Eliminate Possibilities) closes it out: the answer sits just above 3, and only one choice is that small.
Draw the axis of symmetry
The axis of symmetry halves the chord.
Reflecting the circle across the diameter through the arc's midpoint swaps A and B, so that diameter must pass through C and cut the chord in half.
8.G.A.1Draw A DiagramDistance from center to chord
One right triangle gives the centre's distance as 4.
The line from the center to a chord's midpoint is perpendicular to the chord, so half the chord and the center-to-chord distance are the two legs of a right triangle.
The line from the centre to a chord's midpoint meets the chord square on, making a right triangle.
▸ Why?
Folding the circle along that line swaps the chord's two ends, so the line is the chord's perpendicular bisector.
▸ Why?
With that right angle, the radius, half the chord, and the centre-to-chord distance are tied by one equation.
Find the short gap MC
The remaining gap along the axis is just 1.
O, M, and C sit on one line with M between O and C, so the distances simply subtract.
6.NS.C.6Identify SubproblemsCompute AC
A second right triangle gives √(10), choice (A).
AC is the hypotenuse of a small right triangle whose legs are the half-chord AM and the short reach MC.
8.G.B.7Identify SubproblemsDraw the line of symmetry through the center: it splits the chord in half and turns the distance you want into the hypotenuse of a tiny right triangle.
- Draw the axis of symmetry
- Distance from center to chord
- Find the short gap MC
- Compute AC