AMC 10 · 2022 · #14

Grade 11 geometry-2dalgebra
polynomial-factoringquadratic-equationscoordinate-geometrytrigonometric-ratiostangent-addition-formula identify-subproblemsspatial-visualization ↑ Prerequisites: quadratic-equationstrigonometric-ratios
📏 Medium solution 💡 2 insights
Problem
A parabola crosses the horizontal axis at two points and the vertical axis at one. Those three points are the corners of a triangle. Find the tangent of the angle at the corner on the vertical axis.

Pick an answer.

(A)
$\frac{1}{7}$
(B)
$\frac{1}{4}$
(C)
$\frac{3}{7}$
(D)
$\frac{1}{2}$
(E)
$\frac{4}{7}$
How to solve
Strategy Identify Subproblems

There is no formula that reads off the angle between two slanted segments straight from an equation, so the work has to be broken into pieces that each have a formula. Tool #7 (Identify Subproblems) runs the whole solution, and it runs twice. First it says: before any angle can be discussed, the curve has to become three concrete points, so solve for the intercepts. Then it makes the move the problem is really built around — the angle at B is hard, but the y-axis already passes through B and slices that angle into two pieces, and each piece sits in a right triangle whose legs lie along the axes. Two easy angles instead of one hard one. Tool #1 (Draw a Diagram) is what makes that slicing visible and, just as importantly, shows in advance that the angle is sharp rather than wide, which later decides a sign. Tool #15 (Organize Information in More Ways) does the quiet conversion in the middle: the coordinates -5, 3 and -15 have to stop being positions and start being leg lengths 5, 3 and 15 before any trigonometry applies. Tool #4 (Introduce a Variable) finishes it. Neither half-angle is a nameable number of degrees, so give them letters, α and β, keep only their tangents, and let the tangent addition formula put them back together.

1STEP 1

Turn the curve into points

Factoring gives the three points.

x²+2x-15=(x+5)(x-3)=0 ⟹ x=-5 or x=3; x=0 ⟹ y=-15. So A=(-5,0), C=(3,0), B=(0,-15).
2STEP 2

Plot and size up the angle

Two points sit on opposite sides of the axis.

A=(-5,0) left of the y-axis, C=(3,0) right of it, B=(0,-15) on it, 15 units below the origin O=(0,0).
3STEP 3

Cut the angle with the axis

The axis splits it into two right triangles.

∠ ABC=∠ ABO+∠ OBC=α+β, where O=(0,0) and both △ ABO and △ CBO are right-angled at O.
4STEP 4

Read both tangents

Read a tangent off each triangle.

tanα=OA/OB=5/15=1/3, tanβ=OC/OB=3/15=1/5
5STEP 5

Glue with the addition formula

The addition formula gives four sevenths.

tan(∠ ABC)=tan(α+β)=(tanα+tanβ)/(1-tanαtanβ)=(1/3+1/5)/(1-1/3·1/5)=8/15/14/15=8/14=4/7 (E)
Answer
4/7
Put numbers on it. α=arctan1/3≈ 18.43^° and β=arctan1/5≈ 11.31^°, so ∠ ABC≈ 29.74^° — acute, exactly as the sketch predicted, and nowhere near a right angle. Its tangent is tan 29.74^°≈ 0.5714, while 4/7≈ 0.5714. They agree. There is also a check that needs no calculator at all: ∠ ABC is strictly larger than α by itself, and tangent increases as an acute angle grows, so tan(∠ ABC) must exceed tanα=1/3≈ 0.33. That alone eliminates (A) 1/7≈ 0.14 and (B) 1/4=0.25. The trap to watch is (D) 1/2: it is exactly what falls out if the addition formula is written with a plus sign in the denominator, since 8/15/16/15=1/2. That minus sign is not decoration.
💡Key takeaway

When an angle in a coordinate picture is awkward, slide the nearest axis through its corner to cut it into two right-triangle angles, read off each tangent, and glue them back together with the tangent addition formula.

  • Turn the curve into three points
  • Plot the points and size up the angle
  • Let the y-axis cut the angle in two
  • Read each tangent off the legs
  • Glue the angles back with tangent addition