AMC 10 · 2010 · #17
Grade 8 geometry-2dPick an answer.
The area equation hides two separate area computations, so split the work (Tool #7 Identify Subproblems): first find the area of △ ACE, then find the area of the whole hexagon, both in terms of r. Drawing the figure (Tool #1) reveals two structural gifts — the three corner triangles ABC, CDE, EFA are congruent, which forces △ ACE to be equilateral, and the hexagon is a big equilateral triangle with three unit corners sliced off. Since everything is already labelled with the variable r (Tool #4), setting [△ ACE] = 0.7 · [hexagon] collapses to a single quadratic in r, and the question only asks for the sum of its roots.
Triangle ACE is equilateral
The three corner triangles are congruent.
Identical corner triangles must hand back identical long edges, so the inner triangle can only be equilateral.
8.G.A.2Draw A DiagramFind side AC
So the inner triangle is equilateral.
A 120° corner splits into a clean 30-60-90 triangle, so Pythagoras finishes the length with no trig needed.
A hundred twenty degree corner splits into a clean thirty-sixty-ninety triangle, so no trigonometry is needed.
▸ Why?
That triangle's sides sit in a fixed ratio, so the pieces are known the moment the corner is split.
▸ Why?
With the right angle in hand, the remaining length is the hypotenuse over two known legs.
Area of triangle ACE
Its side follows from one right-angle drop.
For an equilateral triangle you only need s², and we already have it as r²+r+1.
6.G.A.1Identify SubproblemsArea of the hexagon
The hexagon is a big triangle minus three corners.
A tidy equiangular hexagon is just a big equilateral triangle with its three tips snipped off.
7.G.B.6Identify SubproblemsSet up and simplify the equation
The share condition becomes a plain quadratic.
The messy √(3)/4 factor is identical on both sides, so the geometry boils down to one clean quadratic.
8.EE.C.7Introduce A VariableSum of all valid r
Its roots sum to 6, choice (E).
You never need the individual roots — the sum of the solutions of r²-6r+1 is just the middle coefficient flipped in sign.
8.EE.C.7Introduce A VariableWrite both areas in terms of r: the inner triangle is equilateral with side² = r²+r+1, and the hexagon is a big triangle minus three unit corners with area factor r²+4r+1. The 70% rule turns into r² - 6r + 1 = 0, and the sum of its two roots is (E) 6.
- Triangle ACE is equilateral
- Find side AC
- Area of triangle ACE
- Area of the hexagon
- Set up and simplify the equation
- Sum of all valid r