AMC 10 · 2004 · #20

Grade 8 geometry-2d

Archetype Coordinate Plane Figure Computation └ Sub-archetype Figure Constraint Equations for Unknown Quantity

equilateral-trianglepythagorean-theoremquadratic-equations convert-to-algebrasymmetry-argument ↑ Prerequisites: pythagorean-theorem

📏 Medium solution 💡 2 insights 📊 Diagram

Problem

Points $E$ and $F$ are located on square $ABCD$ so that $\triangle BEF$ is equilateral. What is the ratio of the area of $\triangle DEF$ to that of $\triangle ABE$ ?

Pick an answer.

(A)

$\frac{4}{3}$

(B)

$\frac{3}{2}$

(C)

$\sqrt{3}$

(D)

(E)

$1+\sqrt{3}$

AMC 10 2004 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

Try it yourself first — the explanation is most useful after you’ve attempted it.

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Toolkit + CCSS Solution

Understand

Restated: A square ABCD has two points E and F placed on its sides so that triangle BEF is equilateral, with E on side AD and F on side DC. Find the ratio of the area of triangle DEF to the area of triangle ABE.

Givens: ABCD is a square.; E lies on side AD and F lies on side DC.; Triangle BEF is equilateral, so its three sides BE, EF, and FB are all equal.

Unknowns: The ratio of the area of triangle DEF to the area of triangle ABE.

Understand

Givens: ABCD is a square.; E lies on side AD and F lies on side DC.; Triangle BEF is equilateral, so its three sides BE, EF, and FB are all equal.

Plan

Primary tool: #4 Introduce a Variable

Secondary: #1 Draw a Diagram, #7 Identify Subproblems

The one length that controls everything is how far E sits from corner A, so name it x. Every other length in the figure then becomes 1 or something built from x, and the equal sides of the equilateral triangle turn into equations through the Pythagorean theorem. The clever part at the end is that the area ratio can be simplified using the equation for x directly, so x never has to be solved for.

Execute — Answer: D

#1 Draw a Diagram 6.EE.B.6 Step 1

Fix the square and name one length

Set the side of the square to 1, which is allowed because only a ratio of areas is wanted.
Let x be the distance AE from corner A up to E on side AD.
Then E is x above A, and since F sits on the top side DC, the leftover piece DE along the left side has length 1 minus x.

$$AB = 1, \qquad AE = x, \qquad DE = 1 - x$$

💡 One unknown length drives the whole picture, so give it a letter and read every other length off from it.

#7 Identify Subproblems 8.G.B.7 Step 2

Match the two side lengths from B

Triangle ABE is right-angled at A, so BE squared equals AB squared plus AE squared, which is 1 plus x squared.
Triangle CBF is right-angled at C, so BF squared equals CB squared plus CF squared, which is 1 plus CF squared.
Because BE and BF are both sides of the equilateral triangle they are equal, forcing CF to equal x.
So F is x to the left of C, and the top piece DF also has length 1 minus x, exactly matching DE.

$$BE^2 = 1 + x^2, \quad BF^2 = 1 + CF^2 \;\Rightarrow\; CF = x \;\Rightarrow\; DF = DE = 1 - x$$

💡 Two equal slant sides leaning on equal square edges must reach equally far in, so the two corner cuts match.

#4 Introduce a Variable 8.G.B.7 Step 3

Set the third side equal too

Triangle DEF is right-angled at D with both legs equal to 1 minus x, so EF squared is 2 times (1 minus x) squared.
Since EF is also a side of the equilateral triangle, it equals BE, whose square is 1 plus x squared.
Setting these two expressions for the same length equal gives one equation in x.

$$EF^2 = 2(1 - x)^2 = 1 + x^2$$

💡 All three sides are the same length, so writing two of them as equal squares pins down x.

#4 Introduce a Variable 8.EE.C.7 Step 4

Simplify to a usable relation

Expand both sides and collect terms.
This turns the equation into x squared minus 4x plus 1 equals 0, which rearranges to x squared plus 1 equals 4x.
Keep it in this form; there is no need to solve for x itself.

$$2 - 4x + 2x^2 = 1 + x^2 \;\Rightarrow\; x^2 - 4x + 1 = 0 \;\Rightarrow\; x^2 + 1 = 4x$$

💡 A tidy relation between x squared and x is more useful here than the messy value of x.

#4 Introduce a Variable 6.G.A.1 Step 5

Compare the two areas

Triangle ABE has legs AB = 1 and AE = x, so its area is one half of x.
Triangle DEF has both legs equal to 1 minus x, so its area is one half of (1 minus x) squared.
The ratio is therefore (1 minus x) squared over x.
Now expand (1 minus x) squared into x squared minus 2x plus 1, group the x squared plus 1 part, and replace it with 4x from the relation: this leaves 4x minus 2x, or 2x.
Dividing by x gives 2, which is choice (D).

$$\frac{[DEF]}{[ABE]} = \frac{\tfrac12(1-x)^2}{\tfrac12 x} = \frac{(1-x)^2}{x} = \frac{(x^2+1) - 2x}{x} = \frac{4x - 2x}{x} = 2$$

💡 Substituting the x-relation into the area expression makes x cancel, leaving a clean number.

[1] #1 6.EE.B.6 Set the side of the square to 1, which is allowed because only a ratio of areas

[2] #7 8.G.B.7 Triangle ABE is right-angled at A, so BE squared equals AB squared plus AE squar

[3] #4 8.G.B.7 Triangle DEF is right-angled at D with both legs equal to 1 minus x, so EF squar

[4] #4 8.EE.C.7 Expand both sides and collect terms. This turns the equation into x squared minu

[5] #4 6.G.A.1 Triangle ABE has legs AB = 1 and AE = x, so its area is one half of x. Triangle

Review

Reasonableness: The relation x squared minus 4x plus 1 equals 0 gives x = 2 minus root 3, about 0.27, which is a believable small distance up the side of a unit square. Plugging back, triangle ABE has area about 0.13 and triangle DEF has area about 0.27, whose quotient is 2, matching the tidy answer. Since DEF spans nearly the full top-left corner while ABE hugs the bottom-left corner, DEF being about twice as large is sensible.

Alternative: Chase angles instead. The three angles at B split the right angle into 15, 60, and 15 degrees, so angle ABE is 15 degrees and AE = tan 15 degrees = 2 minus root 3. Computing the two areas directly from this value and dividing again yields 2.

CCSS standards used (min grade 8)

6.EE.B.6 Use variables to represent numbers and write expressions to solve problems (Naming the distance AE as x and expressing every other length in terms of x.)
8.G.B.7 Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Writing BE, BF, and EF as square roots of leg sums to compare the equal sides of the equilateral triangle.)
8.EE.C.7 Solve linear equations in one variable (Expanding and collecting terms to reduce the side equation to the relation x squared plus 1 equals 4x.)
6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing (Computing the areas of triangles ABE and DEF from their legs and forming their ratio.)

⭐ Name the one unknown length, turn the equal sides into an equation, then feed that equation into the area ratio so the unknown cancels.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Fix the square and name one length

Match the two side lengths from B

Set the third side equal too

Simplify to a usable relation

Compare the two areas

Review

CCSS standards used (min grade 8)

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