AMC 10 · 2012 · #10

Grade 9 geometry-2d
systems-of-equationscircle-equationquadratic-equationsarea-triangles convert-to-algebraidentify-subproblems ↑ Prerequisites: systems-of-equationscoordinate-geometry
📏 Medium solution 💡 3 insights
Problem
Two curves cross, and every crossing point is a corner of one polygon. Find that polygon's area.

Pick an answer.

(A)
24
(B)
27
(C)
36
(D)
37.5
(E)
42
How to solve
Strategy Convert to Algebra

"Points of intersection" is geometry's phrase for "common solutions of a system", so Tool #13 (Convert to Algebra) turns the whole question into two equations in two unknowns. Tool #15 (Organize Information in More Ways) supplies the key move: both equations contain y only as y², and the second one carries 9y² while the first carries y² — so scaling the first equation by 9 makes the two y-terms identical and a single subtraction erases y completely, leaving one quadratic in x. That quadratic proves only necessity: it says which x-values a shared point could have. Tool #3 (Eliminate Possibilities) closes the gap by recovering y for each surviving x and substituting every resulting point back into both original equations — that substitution is what upgrades "can only be among these" into "is exactly these". Tool #1 (Draw a Diagram) then plots the survivors so the base and the height can be read straight off the coordinates, and Tool #7 (Identify Subproblems) keeps the two halves apart: first pin down the points, then compute the area. Worth noticing what this plan never uses: no formula for a circle, no formula for an ellipse, no conic theory at all — just a system of two equations.

1STEP 1

A vertex solves both equations

A corner must satisfy both equations.

x²+y²=25 ; (x-4)²+9y²=81
2STEP 2

Scale, then subtract to erase y

Scaling and subtracting erases one variable.

[(x-4)²+9y²]-9[x²+y²]=81-225 ⟹ (x-4)²-9x²=-144
3STEP 3

Factor the quadratic in x

What remains factors into two candidates.

-8x²-8x+160=0 ⟹ x²+x-20=0 ⟹ (x+5)(x-4)=0 ⟹ x=-5 or x=4
4STEP 4

Recover y and check every candidate

One candidate gives a single point, the other gives two.

x=-5: y²=0 → (-5,0); x=4: y²=9 → (4,3), (4,-3)
5STEP 5

Three points, one triangle

Three points make a triangle.

base=3-(-3)=6, height=4-(-5)=9
6STEP 6

Area of the triangle

Its area is 27, choice (B).

Area=1/2 · 6 · 9=27 ⟹ (B)
Answer
27
Recompute the area a different way as a check. The shoelace formula on (-5,0), (4,3), (4,-3) gives 1/2|(-5)(3-(-3)) + 4((-3)-0) + 4(0-3)| = 1/2 |-30-12-12| = 1/2(54) = 27, matching the base-height computation. Size check: every vertex satisfies x²+y²=25, so the whole triangle sits inside a disk of radius 5 with area 25π ≈ 78.5; an area of 27 fits comfortably inside that and is far from the largest choice, 42. The clean numbers also have a reason worth seeing: (-5,0) is the leftmost point of both curves, (4,3) and (4,-3) are the highest and lowest points of the second curve, and each of the three is exactly 5 from the origin since √(16+9)=5. One count-based warning deserves an answer: two curves of second degree meet in at most four points, so finding only three is a signal to check that none were lost. Nothing was. At (-5,0) the two curves share the same vertical tangent line x=-5 — solving each for x near y=0 gives x=-√(25-y²) and x=4-3√(9-y²), both with derivative 0 at y=0 — so the curves touch there instead of crossing, and that single point absorbs two of the four. That is exactly why the polygon is a triangle and not a quadrilateral, and the answer stands at 27, choice (B).
💡Key takeaway

Intersection points are just the solutions of both equations at once: match the y² terms so one subtraction erases y, solve the quadratic that is left, then plug each candidate back into both equations to prove it is real — the three survivors make a triangle with base 6 and height 9.

  • A vertex solves both equations
  • Scale, then subtract to erase y
  • Factor the quadratic in x
  • Recover y and check every candidate
  • Three points, one triangle
  • Area of the triangle