AMC 10 · 2018 · #25
Grade 10 geometry-2d
Pick an answer.
The picture looks tangled because six objects — three circles, three points — all constrain each other. Tool #1 (Draw a Diagram) cuts it down: add the three centres O₁, O₂, O₃ and the radii to the marked points, and the tangency turns into right angles you can measure with. Tool #17 (Visualize Spatial Relationships) does the one piece of real thinking: deciding which side of line P₁P₂ each centre falls on. That single sign decision is what separates the right answer from a wrong one, because it decides whether two heights add or subtract. Tool #7 (Identify Subproblems) then breaks one side of the triangle into two pieces measured by two different right triangles — a 30-60-90 triangle for the short piece and one Pythagorean step for the long piece. Tool #13 (Convert to Algebra) finishes with the equilateral-area formula and the radical arithmetic, and tool #3 (Eliminate Possibilities) checks the result against the five answers, which also confirms the sign decision made earlier.
Pin down the three centres
All three centre distances are equal.
Two circles touching from outside are as close as they can get without overlapping, so their centres sit exactly one radius plus one radius apart.
10.G-CO.A.1Draw A DiagramTangency becomes a right angle
Tangency becomes a right angle.
At the single point where a line grazes a circle, the radius is the shortest link from centre to line, and the shortest link is always the perpendicular one.
At the point where a line grazes a circle, the radius is the shortest link from centre to line, so it meets it square on.
▸ Why?
A tangent touches at exactly one point, and the radius drawn there is perpendicular to it.
▸ Why?
Any other path from the centre to the line is longer, which is what makes that one the shortest.
Find where O₂ sits
The sixty-degree angle fixes the rest.
The right angle at P₂ is wider than the triangle's 60° corner, so the radius pokes out past the side and lands on the far side of it.
10.G-CO.C.9Visualize Spatial RelationshipsDrop a perpendicular from O₂
Drop a perpendicular from one centre.
A 4-long radius tipped 30° off the line splits into a height of 2 and a shadow of 2√(3) along the line.
10.G-SRT.C.8Identify SubproblemsPythagoras across the two centres
Pythagoras across two centres gives the side.
Two points at known heights on opposite sides of a line are separated by a right triangle whose vertical leg is the sum of the heights.
8.G.B.7Identify SubproblemsArea of the equilateral triangle
Apply the equilateral area formula.
Squaring a sum of two square roots leaves the two squares as whole numbers plus one cross term, which is the only place a new radical can appear.
9.A-SSE.A.2Convert To AlgebraFold the coefficients inside
Folding the coefficients inside gives 552.
A number in front of a square root can be tucked under it by squaring, because k√(n) = √(k² n).
8.EE.A.2Eliminate PossibilitiesA tangent line meets its radius at a right angle, so drop the two centres onto the same side of the triangle: one 30-60-90 triangle hands you 2√(3), one Pythagorean step hands you 2√(7), and the side is their sum.
- Pin down the three centres
- Tangency becomes a right angle
- Find where O₂ sits
- Drop a perpendicular from O₂
- Pythagoras across the two centres
- Area of the equilateral triangle
- Fold the coefficients inside