AMC 10 · 2011 · #20
Grade 10 geometry-2dPick an answer.
Chasing the two circle equations is possible but blind. Tool #16 (Change Focus) is primary: instead of asking "where do these circles cross," ask "is there a point already famous in this triangle that both circles are forced to contain?" The circumcenter O is that point, because D and E are midpoints and O sits on every perpendicular bisector — which manufactures right angles at D, E, F. Tool #1 (Draw a Diagram) makes those right angles visible. Tool #7 (Identify Subproblems) splits the work into two independent halves: prove X = O, then compute the circumradius R. Tool #3 (Eliminate Possibilities) does the real rigor: showing {E, O} is exactly the intersection, so X ≠ E leaves X = O as the only survivor. That last step is where most write-ups skip — finding a common point is necessity; ruling out any third possibility is what makes X = O certain.
Bring in the circumcenter
The circumcenter meets every midpoint squarely.
A point equidistant from two endpoints sits on their perpendicular bisector, and that bisector meets the segment squarely at its midpoint.
A point equally far from two endpoints sits on their perpendicular bisector, which meets the segment at its midpoint.
▸ Why?
Those equidistant points make up exactly the line that folds one endpoint onto the other.
▸ Why?
Every vertex is the same distance from that centre, so all three sit on one circle around it.
Right angles force O onto both circles
Right angles put it on both circles.
A right angle is the fingerprint of a diameter: see a segment at 90° and you are standing on the circle that segment spans.
10.G-C.A.2Change Focus Count The ComplementRule out every other candidate for X
No other point can be the hidden one.
Two different circles can share at most two points, so once you name two of them there is nothing left for a third candidate to be.
10.G-CO.A.1Eliminate PossibilitiesCollapse the sum into three radii
So the sum collapses to three radii.
Every vertex is the same distance from the circumcenter, so a three-term sum shrinks to one measurement tripled.
10.G-C.A.3Identify SubproblemsSplit 13-14-15 with an altitude
An altitude splits the triangle into whole numbers.
One altitude turns an awkward scalene triangle into two right triangles whose sides are familiar Pythagorean triples.
8.G.B.7Identify SubproblemsRead the radius off a similar triangle
Similar triangles give a radius of 65/8.
The altitude and the diameter cut out two right triangles that share an inscribed angle, so their sides are in the same ratio.
10.G-SRT.B.5Identify SubproblemsTriple the radius and match a choice
Tripling gives 195/8, choice (C).
Once the radius is exact, the sum is a single multiplication — and here only an exact value can pick the right choice.
7.NS.A.2Eliminate PossibilitiesMidpoints plus a circumcenter make right angles, and a right angle means you are standing on a circle with that segment as diameter — so both circles are forced through the circumcenter, and the three distances become one radius counted three times.
- Bring in the circumcenter
- Right angles force O onto both circles
- Rule out every other candidate for X
- Collapse the sum into three radii
- Split 13-14-15 with an altitude
- Read the radius off a similar triangle
- Triple the radius and match a choice
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