AMC 10 · 2020 · #12

Grade 9 geometry-2d
Archetype Coordinate Plane Figure Computation └ Sub-archetype Pythagorean / Distance Formula for a Length
chord-perpendicular-from-centerpythagorean-theoremisosceles-right-trianglesymmetry-argument identify-subproblemsconvert-to-algebrasymmetry-argument ↑ Prerequisites: pythagorean-theoremchord-perpendicular-from-center
📏 Medium solution 💡 3 insights
Problem
A circle has radius five root two, and a diameter is drawn. A chord crosses the diameter at a point that sits two root five from one end of the diameter, and the chord meets the diameter at a 45 degree angle. Find the sum of the squares of the two pieces the crossing point cuts the chord into.

Pick an answer.

(A)
96
(B)
98
(C)
$44\sqrt5$
(D)
$70\sqrt2$
(E)
100
How to solve
Strategy Draw a Diagram

Tool #1 (Draw): the one line worth adding to the picture is the perpendicular OM from the center to the chord — it bisects the chord and creates two right triangles. Tool #4 (Variable): name the half-chord x = CM and the offset y = ME instead of chasing the ugly lengths 5√(2) - 2√(5); then CE and DE are just x - y and x + y. Tool #7 (Subproblems): two separate right triangles do all the work — △ OME turns the 45° into OM = ME, and △ OMC turns the radius into x² + y² = 50. Tool #3 (Eliminate): the sum of squares collapses to a single clean number that matches exactly one choice, and the three radical-looking choices are decoys built from the unused length BE.

1STEP 1

Drop a perpendicular

The perpendicular cuts the chord in half.

OM ⊥ CD → CM = MD
2STEP 2

Name the two lengths

The target tidies into a sum of two squares.

CE² + DE² = (x-y)² + (x+y)² = 2x² + 2y²
3STEP 3

What the 45 degrees forces

The 45 degrees makes them equal.

∠ OME = 90°, ∠ OEM = 45° → ∠ MOE = 45° → OM = ME = y
4STEP 4

Pythagoras on the radius triangle

The radius fixes that sum of squares.

x² + y² = OC² = (5√(2))² = 50
5STEP 5

Combine and finish

Doubling gives 100.

CE² + DE² = 2(x² + y²) = 2 · 50 = 100 → (E)
Answer
100
Check the answer with actual numbers. Here OE = OB - BE = 5√(2) - 2√(5) ≈ 2.599, so y = OM = ME = OE · sin 45° ≈ 1.838 and x = √(50 - y²) ≈ 6.828. That makes CE ≈ 4.99 and DE ≈ 8.67, and CE² + DE² ≈ 24.9 + 75.1 = 100. A second, independent check: the Power of a Point at E says CE · DE = AE · EB = (10√(2) - 2√(5))(2√(5)) = 20√(10) - 20 ≈ 43.2, and 4.99 × 8.67 ≈ 43.3 agrees. The whole chord CE + DE ≈ 13.66 is also safely under the diameter 10√(2) ≈ 14.14, as any chord must be. Finally, the general version explains the design of the problem: for a chord meeting the diameter at angle θ, the same algebra gives CE² + DE² = 2r² + 2 · OE² cos 2θ, and θ = 45° kills the second term. The 45° is essential; BE = 2√(5) is a deliberate red herring, which is why three answer choices are dressed up with √(5) and √(2).
💡Key takeaway

Drop a perpendicular from the center to the chord: it splits the chord in half, and the 45° crossing makes the center's distance to the chord equal to the offset of the crossing point from the midpoint. Then CE² + DE² = 2x² + 2y² = 2r² = 100, and the given BE = 2√(5) is never used — a 45° chord always gives twice the square of the radius.

  • Drop a perpendicular to the chord
  • Name the half-chord and the offset
  • The 45 degrees forces OM = ME
  • Pythagoras on the radius triangle
  • Combine and match a choice

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