AMC 10 · 2014 · #10

Grade 8 geometry-2d

Archetype Compound Area by Decomposition / Difference └ Sub-archetype Polygon Triangulation

equilateral-trianglearea-trianglesisosceles-trianglepythagorean-theorem identify-subproblemsconvert-to-algebra ↑ Prerequisites: area-trianglespythagorean-theorem

📏 Medium solution 💡 3 insights

Problem

Three congruent isosceles triangles on the sides of an equilateral one share its total area. Find one of their equal sides.

Pick an answer.

(A)

$\dfrac{\sqrt3}4$

(B)

$\dfrac{\sqrt3}3$

(C)

$\dfrac23$

(D)

$\dfrac{\sqrt2}2$

(E)

$\dfrac{\sqrt3}2$

How to solve

Strategy Draw a Diagram

Nothing here is stated in symbols, so Tool #1 (Draw a Diagram) has to come first: the picture is what tells me each base is a full side of length 1 and each apex sits on the perpendicular bisector of its base. Tool #7 (Identify Subproblems) breaks the work into two easy pieces — the area of the equilateral triangle, then the area of one isosceles triangle — each of which is just 1/2bh once a right triangle is exposed. Tool #4 (Introduce a Variable) names the apex height h so the area condition becomes a one-step equation. Tool #17 (Visualize Spatial Relationships) is what closes the argument at the end: seeing the three triangles fold inward and tile the equilateral triangle shows the required configuration actually exists, instead of only assuming it does.

1STEP 1

Fix the picture

A base and an apex height fix each triangle.

base=1, apex on the perpendicular bisector, at distance h

2STEP 2

Area of the equilateral triangle

The equilateral triangle's area is known.

H=√3/2, [ABC]=1/2 · 1·√3/2=√3/4

3STEP 3

Each triangle gets one third

Each triangle takes exactly a third.

[one isosceles]=1/3·√3/4=√3/12

4STEP 4

Turn the area into a height

That area fixes the apex height.

h/2=√3/12 → h=√3/6

5STEP 5

One use of Pythagoras

One use of Pythagoras gives √3/3.

s²=(1/2)²+(√3/6)²=1/4+1/12=1/3, s=√3/3

6STEP 6

Confirm the configuration is real

The arrangement really exists, choice (B).

r=1/3·√3/2=√3/6=h, R=2/3·√3/2=√3/3 → (B)

Answer

√3/3

Numerically s = √3/3 ≈ 0.577, and two quick bounds confirm that size before any exact work. The equal side is the hypotenuse of a right triangle with one leg 1/2, so s ＞ 1/2 = 0.5, which rules out (A) √3/4 ≈ 0.433. And h = √3/6 ≈ 0.289 ＜ 1/2, so s ＜ √(1/4+1/4) = √2/2 ≈ 0.707, which rules out (D) and (E). Of the two survivors, (C) 2/3 would force h = √(4/9-1/4) = √7/6 ≈ 0.441 and a total area of 3·h/2 ≈ 0.661, far more than the equilateral triangle's √3/4 ≈ 0.433. Only (B) keeps the two areas equal, and a direct recheck agrees: 3·1/2 · 1·√3/6 = √3/4.

💡Key takeaway

Congruent pieces sharing one total each get a third of it, and once a triangle's base and area are known its height is forced — then a single Pythagoras turns that height into the slanted side.

Fix the picture
Area of the equilateral triangle
Each triangle gets one third
Turn the area into a height
One use of Pythagoras
Confirm the configuration is real