AMC 10 · 2008 · #10
Grade 8 geometry-2dPoints A and B are on a circle of radius 5 and AB=6. Point C is the midpoint of the minor arc AB. What is the length of the line segment AC?
Pick an answer.
AMC 10 2008 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 chord $AB$ of length $6$ is drawn in a circle of radius $5$. Point $C$ is the midpoint of the shorter arc between $A$ and $B$. Find the straight-line distance $AC$.
Givens: The circle has radius $5$, so every point on it is $5$ from the center.; $A$ and $B$ lie on the circle and the chord $AB$ has length $6$.; $C$ is the midpoint of the minor (shorter) arc $AB$.
Unknowns: The length of the segment $AC$.
Understand
Restated: A chord $AB$ of length $6$ is drawn in a circle of radius $5$. Point $C$ is the midpoint of the shorter arc between $A$ and $B$. Find the straight-line distance $AC$.
Givens: The circle has radius $5$, so every point on it is $5$ from the center.; $A$ and $B$ lie on the circle and the chord $AB$ has length $6$.; $C$ is the midpoint of the minor (shorter) arc $AB$.
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #3 Eliminate Possibilities
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.
Execute — Answer: A
8.G.A.1 Step 1 Draw the axis of symmetry
- Let $O$ be the center and let $M$ be the midpoint of chord $AB$.
- The whole figure is symmetric across the line through $O$ and $M$: reflecting the circle over this line swaps $A$ and $B$ and folds the minor arc onto itself, so the arc's midpoint $C$ must sit on this same line, and the line crosses $AB$ at a right angle at $M$.
- Thus $O$, $M$, $C$ are three points on one straight line, and $M$ splits $AB$ into $AM=MB=3$.
💡 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.B.7 Step 2 Distance from center to chord
- Look at triangle $OAM$.
- It has a right angle at $M$ because the axis of symmetry meets the chord perpendicularly.
- The side $OA$ is a radius, so $OA=5$, and $AM=3$.
- By the Pythagorean theorem, $OM^2=OA^2-AM^2=25-9=16$, so $OM=4$.
💡 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.
6.NS.C.6 Step 3 Find the short gap MC
- Place the three collinear points on a number line with $O$ at $0$.
- Then $M$ is at $4$, since $OM=4$.
- Point $C$ is on the circle on the same side as $M$, so it is a full radius from the center, $OC=5$, putting $C$ at $5$.
- The gap $MC$ is just the difference: $MC=OC-OM=5-4=1$.
💡 $O$, $M$, and $C$ sit on one line with $M$ between $O$ and $C$, so the distances simply subtract.
8.G.B.7 Step 4 Compute AC
- Now use triangle $AMC$.
- It has a right angle at $M$, since $MC$ runs along the axis of symmetry while $AM$ runs along the chord, and those two directions are perpendicular.
- The legs are $AM=3$ and $MC=1$, so $AC^2=AM^2+MC^2=9+1=10$, giving $AC=\sqrt{10}$.
- That is 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.A.1 Let $O$ be the center and let $M$ be the midpoint of chord $AB$. The whole figur 8.G.B.7 Look at triangle $OAM$. It has a right angle at $M$ because the axis of symmetry 6.NS.C.6 Place the three collinear points on a number line with $O$ at $0$. Then $M$ is a 8.G.B.7 Now use triangle $AMC$. It has a right angle at $M$, since $MC$ runs along the a Review
Reasonableness: The answer $\sqrt{10}\approx3.16$ 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 $\sqrt{10}\approx3.16$, $\tfrac{7}{2}=3.5$, $\sqrt{14}\approx3.74$, $\sqrt{15}\approx3.87$, $4$, only $\sqrt{10}$ is this close to $3$, so (A) is the one that fits.
Alternative: Extend the segment $CM$ across the circle to the far point $D$, so $CD$ is a diameter of length $10$ and $MD=CD-MC=9$. The two chords $AB$ and $CD$ cross at $M$, and the intersecting-chords rule gives $AM\cdot MB=CM\cdot MD$. That is $3\cdot3=CM\cdot 9$, so $CM=1$, matching the number-line result; then $AC=\sqrt{3^2+1^2}=\sqrt{10}$ again.
CCSS standards used (min grade 8)
8.G.A.1Verify experimentally the properties of rotations, reflections, and translations (Using the reflection symmetry across the diameter through $M$ to place $O$, $M$, and $C$ on one line and to bisect the chord.)8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Finding $OM=4$ in right triangle $OAM$ and then $AC=\sqrt{10}$ in right triangle $AMC$.)6.NS.C.6Understand a rational number as a point on the number line (Placing $O$, $M$, $C$ on a number line to get $MC=OC-OM=5-4=1$.)
⭐ 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 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.
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