AMC 10 · 2008 · #10

Grade 8 geometry-2d
chord-perpendicular-from-centerpythagorean-theoremline-symmetry identify-subproblems ↑ Prerequisites: pythagorean-theorem
📏 Medium solution 💡 2 insights
Problem
A circle has radius 5, and its chord AB has length 6. Point C is the midpoint of the shorter arc AB. Find the length of segment AC.

Pick an answer.

(A)
$\ \sqrt{10}$
(B)
$\ \frac{7}{2}$
(C)
$\ \sqrt{14}$
(D)
$\ \sqrt{15}$
(E)
$\ 4$

AMC 10 2008 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Draw a Diagram

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.

1STEP 1

Draw the axis of symmetry

Let O be the center and M the midpoint of AB. The mirror line through O and M also passes through C and meets AB squarely, so AM=MB=3.

AM=MB=6/2=3
2STEP 2

Distance from center to chord

Triangle OAM is right-angled at M, with radius OA=5 and AM=3, so OM²=25-9=16 and OM=4.

OM=√(OA²-AM²)=√(5²-3²)=√(16)=4
3STEP 3

Find the short gap MC

On that line M sits at 4 and C, being on the circle, sits at the radius 5, so the gap is MC=5-4=1.

MC=OC-OM=5-4=1
4STEP 4

Compute AC

Triangle AMC is right-angled at M with legs AM=3 and MC=1, so AC²=9+1=10 and AC=√(10), choice (A).

AC=√(AM²+MC²)=√(3²+1²)=√(10)
Answer
√(10)
The answer √(10)≈3.16 is just a bit more than the half-chord AM=3, which is exactly right: C pokes only MC=1 off the chord, so AC should be slightly longer than AM. It is also far under the diameter 10, as any single segment must be. Scanning the choices √(10)≈3.16, 7/2=3.5, √(14)≈3.74, √(15)≈3.87, 4, only √(10) is this close to 3, so (A) is the one that fits.
💡Key takeaway

Draw 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