AMC 10 · 2012 · #16
Grade 10 geometry-2dPick an answer.
Only one fact links the unknown radius to the three given lengths: OX and OY are radii of C₁ and, at the same time, equal chords of C₂. Seen from Z, equal chords subtend equal angles, so line ZO bisects angle XZY. A bisector is a mirror, and mirroring Y across it drops the length 7 onto segment ZX, which converts the pair ZX, ZY into the power of Z with respect to C₁. That single quantity also equals OZ² - r², and the radius falls out. Drawing the figure correctly comes first, because which arc of C₂ holds Z decides whether the equal-angle step is true at all.
Put all four points on one circle
All four points lie on one circle.
A radius of one circle shows up as a chord of the other, which is how a length belonging to C₁ can be read off C₂.
10.G-C.A.2Draw A DiagramPin down which arc holds Z
The point's arc is pinned down by the distances.
The word "exterior" in the problem is not decoration; it is exactly what fixes the order of the four points around C₂.
10.G-CO.A.1Visualize Spatial RelationshipsShow ZO bisects angle XZY
Equal arcs mean the angle is bisected.
Two equal chords look equally wide from any point on the far side, so Z sees OX and OY at the same angle.
Two equal chords look equally wide from any point on the far side, so the centre line bisects that angle.
▸ Why?
Both chords are radii of the same circle, so they have exactly the same length.
▸ Why?
Equal lengths sit across from equal angles, so the two viewing angles must match.
Fold Y across the bisector
Folding across that bisector builds a new point.
The bisector acts as a mirror, and the mirror turns the loose length ZY into a piece of the segment ZX.
10.G-SRT.B.5Visualize Spatial RelationshipsSwitch to the power of Z
The power of the outside point closes the argument.
Every line drawn through Z reports the same "distance to the circle", so an easy line prices the hard one.
10.G-SRT.B.5Change Focus Count The ComplementSolve for the radius
Solving gives √30, choice (D).
ZX · ZY is the power of Z in disguise, so it subtracts straight off OZ².
8.EE.A.2Introduce A VariableCheck the figure really exists
Coordinates confirm the figure exists.
A value forced by the conditions is only the answer once you can point at a figure that actually has it.
10.G-GPE.A.1Work BackwardsThe two radii OX and OY are equal chords of the big circle, so Z sees them at equal angles and ZO splits angle XZY in half; mirroring across that bisector turns ZX · ZY into the power of Z, giving r² = OZ² - ZX · ZY.
- Put all four points on one circle
- Pin down which arc holds Z
- Show ZO bisects angle XZY
- Fold Y across the bisector
- Switch to the power of Z
- Solve for the radius
- Check the figure really exists
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Same archetype — closest grade level first.
