AMC 10 · 2005 · #15

Grade 8 geometry-2d

Archetype Coordinate Plane Figure Computation └ Sub-archetype Axis-Aligned Figure Area from Coordinates

coordinate-geometryarea-trianglespythagorean-theorem convert-to-algebraidentify-subproblems ↑ Prerequisites: coordinate-geometryarea-trianglespythagorean-theorem

📏 Long solution 💡 3 insights 📊 Diagram

Problem

A point on a diameter divides it so that one piece is twice the other, and a perpendicular there meets the circle. That meeting point is one end of a second diameter. Find the ratio of two triangle areas in the figure.

Pick an answer.

(A)

$\frac{1}{6}$

(B)

$\frac{1}{4}$

(C)

$\frac{1}{3}$

(D)

$\frac{1}{2}$

(E)

$\frac{2}{3}$

How to solve

Strategy Draw a Diagram

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.

1STEP 1

Fix the radius and locate C

Scaling is free, so set the radius to one and place the centre at the origin.

AC:CB=1:2 → AC=1/3 · 2=2/3, C=(-1/3,0)

2STEP 2

Find the height DC with the circle

The circle's equation gives the height of the perpendicular.

(-1/3)²+y²=1 → y²=8/9 → DC=2√(2)/3

3STEP 3

Locate E opposite D

The far end of the second diameter is the reflection through the centre.

E=-D=(1/3,-2√(2)/3), horizontal gap to line DC=2/3

4STEP 4

Area of triangle ABD

One triangle uses the diameter as base, giving a clean area.

[△ ABD]=1/2 · 2·2√(2)/3=2√(2)/3

5STEP 5

Area of triangle DCE

The other uses the vertical segment as base, with a horizontal height.

[△ DCE]=1/2·2√(2)/3·2/3=2√(2)/9

6STEP 6

Divide to get the ratio

Dividing cancels the roots, giving 1/3, choice (C).

2√(2)/9/2√(2)/3=1/9/1/3=1/3 → (C)

Answer

1/3

The 2√(2) common factor cancelling is a strong sign the setup is right: a clean ratio like 1/3 is exactly what a well-posed multiple-choice answer should look like. A second check comes from the shared segment DC: both triangles can be seen as having height DC, so their ratio is the ratio of the matching bases. For △ ABD the base is AB=2; for △ DCE the relevant width is E's horizontal reach 2/3, giving 2/3/2=1/3 — the same answer, so (C) is consistent.

💡Key takeaway

Drop 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