AMC 10 · 2004 · #21

Grade 11 algebrageometry-2d
Archetype Coordinate Plane Figure Computation └ Sub-archetype Figure Constraint Equations for Unknown Quantity
quadratic-equationsvieta-formulasoptimization convert-to-algebraextreme-principle ↑ Prerequisites: quadratic-equations
📏 Medium solution 💡 4 insights
Problem
An ellipse sits in the first quadrant. As a point runs over it, the ratio of the second coordinate to the first has a largest and a smallest value. Find the sum of those two values.

Pick an answer.

(A)
3
(B)
$\sqrt{10}$
(C)
$\frac 72$
(D)
$\frac 92$
(E)
$2\sqrt{14}$
How to solve
Strategy Introduce a Variable

Tool #16 (Change Focus) makes the problem tractable: instead of hunting for the two extreme points, ask which slopes m occur at all, since y/x is exactly the slope of the line from the origin through the point. Tool #4 (Introduce a Variable) names that slope m and substitutes y=mx, turning 'does the line y=mx hit the ellipse?' into 'does a quadratic in x have a real root?'. Tool #14 (Extreme Principle) then reads off the boundary: the achievable slopes form a closed interval whose endpoints are exactly where the quadratic's discriminant hits 0, so a and b are the two roots of one quadratic in m and Vieta hands over their sum without ever computing them. Tool #16 returns at the end for that last move: the question asks only for a sum, which is cheaper than either root.

1STEP 1

Read y/x as a slope

The ratio is a slope through the origin, and the ellipse misses the vertical axis.

x=0 → 3y²-20y+40=0, discriminant = -80 ＜ 0
2STEP 2

Substitute y=mx

Substituting that line gives an honest quadratic for every slope.

(3m²+m+2)x² - (20m+11)x + 40 = 0, 3m²+m+2 ＞ 0 for all m
3STEP 3

Slope m occurs exactly when the discriminant is not negative

A slope occurs exactly when its discriminant is not negative.

D(m) = (20m+11)² - 160(3m²+m+2) = -80m² + 280m - 199
4STEP 4

The achievable slopes form a closed interval

So the achievable slopes form a closed interval whose ends are two roots.

80m²-280m+199=0, 280²-4 · 80 · 199 = 14720 ＞ 0
5STEP 5

Add the roots with Vieta

Vieta adds those roots without computing either, giving 7/2, choice (C).

a+b = 280/80 = 7/2 (C)
Answer
7/2
Solving 80m²-280m+199=0 gives m=35±√(230)/20, so b≈ 0.992 and a≈ 2.508. Both are positive, which is required since the ellipse lies in the first quadrant, and their sum is 70/20=7/2=3.5, matching choice (C). The interval also passes spot checks: D(2) = -320+560-199 = 41 ＞ 0, so slope 2 really is attained, while D(0) = -199 ＜ 0 and D(3) = -720+840-199 = -79 ＜ 0, so slopes 0 and 3 are not. Finally, the endpoint slope b gives the touching point x=(20b+11)/(2(3b²+b+2))≈ 2.594, y≈ 2.573, which does satisfy the original equation to within rounding and lies in the first quadrant, so the extremes are genuinely attained.
💡Key takeaway

y/x is the slope of a line through the origin, so plug y=mx in and ask when the resulting quadratic still has a solution: the discriminant -80m²+280m-199 is non-negative exactly between its two roots, and those roots add to 280/80=7/2.

  • Read y/x as a slope
  • Substitute y=mx
  • Slope m occurs exactly when the discriminant is not negative
  • The achievable slopes form a closed interval
  • Add the roots with Vieta

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