AMC 10 · 2021 · #13

Grade 10 geometry-2d

Archetype Coordinate Plane Figure Computation └ Sub-archetype Figure Constraint Equations for Unknown Quantity

angle-bisector-theoremcoordinate-geometryslope-interceptgolden-ratio convert-to-algebraidentify-subproblems ↑ Prerequisites: angle-bisector-theoremcoordinate-geometry

📏 Long solution 💡 3 insights

Problem

Two lines of slope one and slope three both pass through the origin, making an acute angle there. The ray that cuts that acute angle into two equal halves also passes through the origin, so it has its own slope. Find that slope.

Pick an answer.

(A)

$\frac{1+\sqrt{5}}{2}$

(B)

$\frac{1+\sqrt{7}}{2}$

(C)

$\frac{2+\sqrt{3}}{2}$

(D)

(E)

$\frac{2+\sqrt{5}}{2}$

How to solve

Strategy Draw a Diagram

An angle by itself is hard to compute with, but a triangle is not. Since all three lines run through the origin, I can cut across all of them with the single vertical line x = 1 and lose nothing. That one stroke turns the picture into a triangle whose two sides at the origin I can measure with the distance formula, and it turns the unknown slope into the height of a point on the opposite side. The Angle Bisector Theorem then converts "equal angles" into a plain ratio of lengths, and the rest is one linear equation.

1STEP 1

Read every slope off the line x = 1

Read every slope off one vertical line.

P = (1, 1), Q = (1, 3), R = (1, k), 1 ＜ k ＜ 3

2STEP 2

Turn the angle into a triangle

Turn the angle into a triangle.

OP = √(1² + 1²) = √(2), OQ = √(1² + 3²) = √(10), PQ = 3 - 1 = 2

3STEP 3

Equal angles become a ratio of lengths

The bisector splits the far side in the sides' ratio.

PR/RQ = OP/OQ = √(2)/√(10) = √(2/10) = √(1/5) = 1/√(5)

4STEP 4

Split a segment of length 2

Split the segment of length two by that ratio.

PR + RQ = 2, RQ = √(5) PR ⟹ PR(1 + √(5)) = 2 ⟹ PR = 2/(1 + √(5))

5STEP 5

Rewrite the length without a radical below

Clear the radical from the denominator.

PR = 2/(1 + √(5)) · (√(5) - 1)/(√(5) - 1) = (2(√(5) - 1))/(5 - 1) = (2(√(5) - 1))/4 = (√(5) - 1)/2

6STEP 6

Climb from P up to R

Climbing up gives the golden ratio.

k = 1 + PR = 1 + (√(5) - 1)/2 = (2 + √(5) - 1)/2 = (1 + √(5))/2

Answer

(1+√(5))/2

The value (1+√(5))/2 ≈ 1.618 lies strictly between the two given slopes 1 and 3, as any line inside the acute angle must. It also sits closer to 1 than to 3, which is exactly what the Angle Bisector Theorem predicted: the shorter side OP = √(2) pulls the crossing point toward P, in the ratio 1 : √(5). An independent angle check confirms it: y=x makes 45° with the x-axis and y=3x makes about 71.57°, so the bisector must sit at about 58.28°, and tan 58.28° ≈ 1.618. This also exposes the trap in choice (D): arctan 2 ≈ 63.43° is not halfway between 45° and 71.57°, so averaging slopes gives the wrong line. As a last check, k = (1+√(5))/2 satisfies k² = k + 1, since 1.618² ≈ 2.618.

💡Key takeaway

Cutting both lines with the single vertical line x = 1 turns a slope question into a length question, and the angle bisector splits that segment in the same ratio as the two sides it sits between.

Read every slope off the line x = 1
Turn the angle into a triangle
Equal angles become a ratio of lengths
Split a segment of length 2
Rewrite the length without a radical below
Climb from P up to R