AMC 10 · 2021 · #21

Grade 11 geometry-2d
ellipse-eccentricitycomplex-numberscoordinate-geometrysymmetry-argument convert-to-algebraidentify-subproblems ↑ Prerequisites: ellipse-eccentricity
📏 Long solution 💡 3 insights
Problem
A fifth-degree equation is handed to us already factored. Each of its five solutions is plotted as a point in the plane, and exactly one ellipse passes through all five. Its eccentricity is the square root of a fraction of two coprime positive integers. Find the sum of those two integers.

Pick an answer.

(A)
7
(B)
9
(C)
11
(D)
13
(E)
15
How to solve
Strategy Introduce a Variable

A general conic carries five unknown coefficients, which is exactly as many as we have points — solvable, but ugly. Tool #1 (Draw a Diagram) cuts that down first: plotting the five roots shows they come in mirror pairs across the x-axis, and since the ellipse through them is unique, the ellipse must have that same mirror symmetry. That single observation kills the xy and y terms and puts the center on the x-axis. Tool #4 (Introduce a Variable) then names what is left — the center (h,0) and the two squared axis lengths a² and b² — so three unknowns absorb all five points. Tool #7 (Identify Subproblems) keeps the finish orderly: solve the quadratics first, then h, then a² and b², then the eccentricity, one small job at a time.

1STEP 1

Solve for the five roots

Read the roots off each factor.

z-1=0→ z=1; (z+1)²=-3→ z=-1± i√3; (z+2)²=-2→ z=-2± i√2
2STEP 2

Read the symmetry

The points are symmetric about the axis.

(1,0), (-1,±√3), (-2,±√2) ⟶ ((x-h)²)/a²+y²/b²=1
3STEP 3

Use the point on the axis

That point fixes one axis.

((1-h)²)/a²+0/b²=1 ⟹ a²=(1-h)²
4STEP 4

Find the centre

The other points fix the centre.

b²=3a²/(a²-(1+h)²)=2a²/(a²-(2+h)²) → a²=3(2+h)²-2(1+h)²=h²+8h+10, and h²+8h+10=(1-h)²=h²-2h+1 → 10h=-9, h=-9/10
5STEP 5

Read both axis lengths

Compute both axis lengths.

a²=(19/10)²=361/100, b²=(3·361/100)/(361/100-1/100)=(3 · 361)/360=361/120
6STEP 6

Convert to eccentricity

Converting and adding gives 7.

e²=1-b²/a²=1-5/6=1/6 → e=√(1/6), m+n=1+6=7 (A)
Answer
7
Check every point against ((x+9/10)²)/361/100+y²/361/120=1. For (1,0): (19/10)²/361/100=1. For (-1,±√3): 1/100/361/100+3/361/120=1/361+360/361=1. For (-2,±√2): 121/100/361/100+2/361/120=121/361+240/361=1. All five points check out. The shape is also believable: the center is (-9/10,0) with a=19/10=1.9 and b=√(361/120)≈ 1.735, so the ellipse is only slightly wider than tall, and e≈ 0.408 is small — exactly what a nearly round ellipse should give. Since 1 and 6 share no factor, √(1/6) is already in the required form and m+n=7, choice (A).
💡Key takeaway

When the points you must fit are mirror images of each other and only one curve fits them, the curve is forced to be a mirror image too — that free symmetry is what turns five unknowns into one.

  • Solve for the five roots
  • Plot the points and read the symmetry
  • Use the point on the axis
  • Eliminate the axis lengths to find h
  • Read off both axis lengths
  • Convert axes into eccentricity