AMC 10 · 2015 · #12

Grade 9 geometry-2d

Archetype Coordinate Plane Figure Computation └ Sub-archetype Axis-Aligned Figure Area from Coordinates

quadratic-equationscoordinate-geometryarea-triangles convert-to-algebrasymmetry-argument ↑ Prerequisites: quadratic-equationscoordinate-geometry

📏 Long solution 💡 3 insights

Problem

Two curves meet the axes at exactly four points that form a kite of known area. Find the sum of the two coefficients.

Pick an answer.

(A)

(B)

1.5

(C)

(D)

2.5

(E)

How to solve

Strategy Eliminate Possibilities

Nothing in this problem can be computed until the picture is pinned down, so the plan starts with counting, not with area. Tool #2 (Make a Systematic List) writes down every point where each curve can touch an axis, as a function of the signs of a and b. Tool #3 (Eliminate Possibilities) then uses the phrase "exactly four points" to kill every configuration but one — this is the step the problem really turns on, and it deserves a proof rather than the word "clearly," because a careless reading leaves configurations alive in which a+b has no single value. Tool #1 (Draw a Diagram) plots the surviving four points and shows that the kite condition is automatic, so the only quantitative information left is the area. Tool #7 (Identify Subproblems) gets that area by cutting the kite into two triangles instead of quoting a memorised diagonal formula, and Tool #11 (Work Backwards) runs the intercepts back into the two equations to recover a and b.

1STEP 1

List every axis crossing

Each curve crosses both axes.

x=0: (0,-2) and (0,4). y=0: x²=2/a and x²=4/b

2STEP 2

Count forces shared x-intercepts

Having only four forces a shared pair.

2/a=4/b=s², s ＞ 0, a ＞ 0, b ＞ 0

3STEP 3

Plot the points and check the kite

Plotting shows the shape really is a kite.

(0,4), (s,0), (0,-2), (-s,0); upper sides √(s²+16), lower sides √(s²+4)

4STEP 4

Cut the kite into two triangles

It cuts into two easy triangles.

[kite]=1/2(2s)(4)+1/2(2s)(2)=4s+2s=6s

5STEP 5

Use the area to get the width

The area fixes the shared crossing at 2.

6s=12 → s=2, so the shared x-intercepts are (± 2,0)

6STEP 6

Read off a and b

Reading both coefficients gives 3/2, choice (B).

4a-2=0→ a=1/2; 4-4b=0→ b=1; a+b=3/2 → (B)

Answer

1.5

Put the numbers back. With a=1/2 and b=1 the curves are y=1/2x²-2 and y=4-x². The first meets the axes at (0,-2) and (± 2,0); the second at (0,4) and (± 2,0). The union is {(0,-2),(0,4),(2,0),(-2,0)} — exactly four points, as required, and not five or six. Their side lengths are √(20),√(8),√(8),√(20): two pairs of adjacent equal sides and no rhombus, so it is a kite, and the area is 4(2)+2(2)=12. A size check kills the other choices without solving anything: the shared intercepts satisfy s²=2/a with 2/a=4/b, which forces s²=6/(a+b), so a larger a+b means a narrower and hence smaller kite. Choice (A) a+b=1 gives s=√(6) and area 6√(6)≈ 14.7, too big; choice (E) a+b=3 gives s=√(2) and area 6√(2)≈ 8.5, too small. The true value has to lie between 1 and 3 and closer to the small end, which is where 1.5 sits.

💡Key takeaway

Before computing anything, use the word "exactly" to pin down the picture — here it forces both parabolas to hit the x-axis at the same two points, and after that the kite is just two triangles stacked on a shared base.

List every axis crossing
Count forces shared x-intercepts
Plot the points and check the kite
Cut the kite into two triangles
Use the area to get the width
Read off a and b