AMC 10 · 2014 · #21

Grade 10 geometry-2d
coordinate-geometrysimilar-trianglesthirty-sixty-ninety-trianglepythagorean-theoremquadratic-equations convert-to-algebracoordinate-geometry ↑ Prerequisites: similar-trianglespythagorean-theorem
📏 Long solution 💡 5 insights 📊 Diagram
Problem
Two congruent rectangles sit inside a square, one square-on and one tilted. Find one segment's length.

Pick an answer.

(A)
$\frac{1}{2}(\sqrt{6}-2)$
(B)
$\frac{1}{4}$
(C)
$2-\sqrt{3}$
(D)
$\frac{\sqrt{3}}{6}$
(E)
$1-\frac{\sqrt{2}}{2}$
How to solve
Strategy Introduce a Variable

The picture hands over no length at all, so nothing can be measured or chased directly. The real difficulty is that there are two hidden unknowns, not one: the width BE and the tilt of JKHG. Tool #4 (Introduce a Variable) names both of them up front — x=BE for the width, and a unit direction vector (u,v) for the tilt. Tool #1 (Draw a Diagram) and Tool #17 (Visualize Spatial Relationships) do the work of turning "JKHG is a rectangle" into coordinates: put the square on axes, then build the whole rectangle from J, the direction (u,v), and its quarter turn (-v,u), so being a rectangle costs no equations at all. Tool #13 (Convert to Algebra) then writes down the only facts the figure actually supplies — which corner is stuck to which line — as three plain equations. It is worth being strict here: a tempting shortcut is to stare at the picture, declare the four corner triangles similar and read off equal segments, but every one of those claims is a figure-reading, not a proof. The incidence conditions are the whole content, and they are enough. Tool #7 (Identify Subproblems) extracts the tilt u twice — once from the requirement that JG has length 1, once from the requirement that K lands on EF — and the two expressions must agree, which is the equation that decides everything. Tool #3 (Eliminate Possibilities) discards the two solutions that the algebra produces but the geometry forbids: the untilted case and the root bigger than 1.

1STEP 1

Put the square on axes

Coordinates name every corner.

B=(0,0), A=(1,0), C=(0,1), D=(1,1)
2STEP 2

Name BE and cash in the congruence

The congruence transfers both dimensions.

x=BE, E=(x,0), F=(x,1), JG=1, JK=x
3STEP 3

Build the tilted rectangle from one direction

One direction builds the tilted rectangle.

J=(j,0), G=(j+u, v), K=(j-xv, xu), H=(j-xv+u, xu+v), u²+v²=1
4STEP 4

Write the incidence facts as equations

Each contact point is one equation.

j+u=1, xu+v=1, j-xv=x, u²+v²=1
5STEP 5

Get the tilt from the length condition

The length condition gives the tilt once.

u²+(1-xu)²=1 ⟹ u[(1+x²)u-2x]=0 ⟹ u=2x/(1+x²)
6STEP 6

Get the tilt a second time from K

The contact condition gives it again.

(1-u)-x(1-xu)=x ⟹ 1-2x=u(1-x²) ⟹ u=(1-2x)/(1-x²)
7STEP 7

Make the two tilts agree

Matching them gives a plain quadratic.

2x/(1+x²)=(1-2x)/(1-x²) ⟹ x²-4x+1=0 ⟹ x=2±√(3), and 0 < x < 1 forces x=2-√(3)
8STEP 8

Rebuild the figure and confirm

Rebuilding confirms 2-√3, choice (C).

u=1/2, v=√(3)/2, j=1/2 ⟹ BE=2-√(3) → (C)
Answer
2-√(3)
First a size check: 2-√(3)≈ 0.268, comfortably between 0 and 1/2, so EBCF is a narrow rectangle hugging side BC — which is what the figure shows. This is a problem where estimating cannot substitute for the algebra: the choices are ≈ 0.225, 0.25, 0.268, 0.289, 0.293, all crowded into one narrow band. Next an area audit, using pieces the derivation never used. With j=1/2, u=1/2, v=√(3)/2 the square splits into the two rectangles plus four right triangles at A, D, F and E. The two rectangles contribute 2x=4-2√(3). Triangles JAG and HFK each have legs 1/2 and √(3)/2, contributing 2·√(3)/8=√(3)/4. Triangles GDH and KEJ each have legs x·1/2 and x·√(3)/2, contributing x²·√(3)/4=(7√(3)-12)/4. The total is (4-2√(3))+√(3)/4+(7√(3)-12)/4=4-2√(3)+2√(3)-3=1, exactly the area of the square, so the six pieces tile it with nothing left over. Two more consistency notes. The rejected root is no accident: (2-√(3))(2+√(3))=1, so the quadratic's roots are reciprocals, and only the one below 1 can be a length inside a unit square. And the reading of the figure taken in step 2 does not matter — if instead JG=x and JK=1, the same construction gives j+xu=1, u+xv=1 and j-v=x, which reduce to v=2x/(1+x²) and v=(1-2x)/(1-x²), the identical pair of equations with v in place of u, hence the identical quadratic and the identical answer (C).
💡Key takeaway

A tilted shape has two secrets — where it starts and how far it leans — so name both, write down which corner is forced onto which side, and the picture turns into one quadratic that answers itself.

  • Put the square on axes
  • Name BE and cash in the congruence
  • Build the tilted rectangle from one direction
  • Write the incidence facts as equations
  • Get the tilt from the length condition
  • Get the tilt a second time from K
  • Make the two tilts agree
  • Rebuild the figure and confirm