AMC 10 · 2013 · #24

Grade 11 geometry-2d

angle-bisector-theoremsimilar-triangleslaw-of-cosinesmedian-of-triangleequilateral-triangle double-countingidentify-subproblemswork-backwards ↑ Prerequisites: similar-triangleslaw-of-cosines

📏 Long solution 💡 4 insights

Problem

A median and a bisector cross so that a small triangle comes out equilateral. Find a squared length.

Pick an answer.

(A)

$\frac{10-6\sqrt{2}}{7}$

(B)

$\frac{2}{9}$

(C)

$\frac{5\sqrt{2}-3\sqrt{3}}{8}$

(D)

$\frac{\sqrt{2}}{6}$

(E)

$\frac{3\sqrt{3}-4}{5}$

How to solve

Strategy Organize Information in More Ways

The picture hands over more angles than lengths, so start by chasing angles: the two 60° angles of the equilateral triangle sit at crossings of straight lines, and supplements spread them everywhere. The key is then that the angle bisector at C makes the two sides of angle C interchangeable, so each angle found on one side of the bisector pairs with an equal angle on the other side. That produces two different pairs of similar triangles, and both pairs measure the same ratio CX/CN. Writing one quantity two ways and setting the two expressions equal is what pins down BC; after that a single Law of Cosines finishes the job.

1STEP 1

Spread the two 60-degree angles

The equilateral corner spreads known angles.

∠ BXC = 120°, ∠ MXC = 60°, ∠ BNC = 60°, ∠ ANC = 120°

2STEP 2

Pair the 60-degree angles across the bisector

One pairing gives a pair of similar triangles.

△ CXM ∼ △ CNB → CX/CN = CM/CB = 1/BC

3STEP 3

Pair the 120-degree angles the same way

The other pairing gives another pair.

△ CXB ∼ △ CNA → CX/CN = CB/CA = BC/2

4STEP 4

One ratio, computed two ways

One ratio computed twice fixes a side.

1/BC = CX/CN = BC/2 → BC² = 2 → BC = √(2)

5STEP 5

Turn the ratio into a length

The same ratio turns into a length.

XN = CN - CX = (√(2)-1) CX = BX → CX = BX/(√(2)-1) = (√(2)+1) BX

6STEP 6

Law of Cosines closes triangle BXC

The law of cosines closes it, choice (E).

2 = BC² = b²(1 + (√(2)+1)² + (√(2)+1)) = (5+3√(2)) b² → BX² = 2/(5+3√(2)) = (10-6√(2))/7

Answer

(10-6√(2))/7

Every step above only says what must be true if such a triangle exists, so the configuration deserves an existence check rather than just an internal consistency check. Build it forward instead: take AC = 2, BC = √(2), and AB = BN + NA = (1+√(2))BX with BX = √((10-6√(2))/7) ≈ 0.46518, so AB ≈ 1.12303. These satisfy the triangle inequality, since 1.41421 + 1.12303 ＞ 2. Placing C = (0,0) and A = (2,0) puts B ≈ (1.18463, 0.77243); then M = (1,0), the point N with AN/NB = CA/CB is N ≈ (1.52237, 0.45247), and BM ∩ CN is X ≈ (1.07650, 0.31991). Measuring the three sides of △ BXN gives BX = XN = NB ≈ 0.465176: the triangle really is equilateral, CN really bisects angle C into two angles of 16.55°, and X really lies inside both segments. So a triangle with all the stated properties exists and the answer is not vacuous. Two extra spot checks agree: the median formula gives BM² = (2AB² + 2BC² - AC²)/4 ≈ 0.63069, matching BM = BX + XM ≈ 0.79417, and BX/XM = √(2) = BC/CM as the bisector of angle BCM inside triangle BCM demands. Numerically (10-6√(2))/7 ≈ 0.2164 is the smallest of the five choices, with the other four packed between 0.222 and 0.240, which is why only an exact computation decides the problem.

💡Key takeaway

An angle bisector makes the two arms of an angle interchangeable, so every matched angle gives a pair of similar triangles; measure the same ratio with two different pairs, set the two answers equal, and the triangle has nowhere left to hide.

Spread the two 60-degree angles
Pair the 60-degree angles across the bisector
Pair the 120-degree angles the same way
One ratio, computed two ways
Turn the ratio into a length
Law of Cosines closes triangle BXC