AMC 10 · 2005 · #23
Grade 8 geometry-2d
Pick an answer.
AMC 10 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
A ratio of areas does not depend on the circle's size, so Tool #4 (Introduce a Variable) lets us pin the radius to 1 and drop a coordinate grid on the picture. Tool #1 (Draw a Diagram) then turns every named point into a pair of coordinates: once A, B, C, D, E are located, both triangles reduce to simple base-times-height calculations. Tool #7 (Identify Subproblems) splits the work into three clean sub-tasks — find C from the ratio, find the height DC from the circle, and find E as the point opposite D — after which each area is one line. Tool #3 (Eliminate Possibilities) matches the exact fraction to the five choices at the end.
Fix the radius and locate C
Scaling keeps a ratio, so take radius 1: A=(-1,0), B=(1,0). AC:CB = 1:2 makes AC=2/3, so C=(-1/3, 0).
A ratio ignores size, so we are free to pick the roundest radius and read positions straight off the axis.
6.RP.A.3Introduce A VariableFind the height DC with the circle
D sits straight above C, so x=-1/3 in x²+y²=1 gives y²=8/9: the vertical segment DC has length 2√(2)/3.
A point on a circle obeys the Pythagorean relation between its coordinates and the radius.
8.G.B.7Draw A DiagramLocate E opposite D
DE is a diameter, so E flips both of D's signs: E=(1/3, -2√(2)/3), sitting a sideways 2/3 from line DC.
The two ends of a diameter are mirror images through the center, so one is the other with both signs flipped.
The two ends of a diameter are mirror images through the centre, so one is the other with both signs flipped.
▸ Why?
Both ends sit one radius from the centre, so the centre is exactly halfway between them.
▸ Why?
A half turn about the centre moves the point without stretching, landing it on the opposite end.
Area of triangle ABD
Base AB has length 2 and D's height above it is its y-coordinate 2√(2)/3, so [△ ABD]=1/2 · 2·2√(2)/3=2√(2)/3.
With the base flat on the axis, the apex's height is simply how high up it sits.
6.G.A.1Identify SubproblemsArea of triangle DCE
Base DC is 2√(2)/3 and E's sideways reach 2/3 is the matching height, so [△ DCE]=1/2·2√(2)/3·2/3=2√(2)/9.
For a vertical base, the far vertex's 'height' is just how far sideways it reaches.
6.G.A.1Identify SubproblemsDivide to get the ratio
Dividing cancels the common 2√(2): 2√(2)/9 ÷ 2√(2)/3 = 1/3, which is none of 1/6, 1/4, 1/2, 2/3 — so choice (C).
The messy 2√(2) appears in both areas, so it cancels and only the tidy denominators decide the ratio.
6.NS.A.1Eliminate PossibilitiesDrop the figure on a grid with radius 1, read off every point, and both triangles become base-times-height; the ugly √(2) cancels and the ratio is just 1/3.
- Fix the radius and locate C
- Find the height DC with the circle
- Locate E opposite D
- Area of triangle ABD
- Area of triangle DCE
- Divide to get the ratio