AMC 10 · 2024 · #20

Grade 10 geometry-2d

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

median-of-trianglepolygon-inequalityoptimizationarea-triangles extreme-principleidentify-subproblems ↑ Prerequisites: area-trianglespolygon-inequality

📏 Medium solution 💡 3 insights

Problem

Points A, B, C lie in the plane with AB = 40 and AC = 42. Let x be the distance from A to the midpoint of BC, and let f(x) be the area of △ ABC. The set of x for which such a triangle exists is an open interval (p, q), and the largest area r is reached at x = s. Find p + q + r + s.

Pick an answer.

(A)

909

(B)

910

(C)

911

(D)

912

(E)

913

How to solve

Strategy Draw a Diagram

The median x is awkward because the triangle inequality talks about sides, not medians. Tool #1 (Draw a Diagram) fixes that with one extra line: extend the median past the midpoint until it doubles in length. That single stroke turns the median into the third side of a brand-new triangle whose other two sides are exactly 40 and 42, so the ordinary triangle inequality applies directly and hands over both endpoints of the domain. Tool #14 (Extreme Principle) then does the other half of the problem twice: it reads the domain's open endpoints as the two flattened, zero-area positions, and it finds the largest area by pushing the height from C to line AB as high as it can possibly go. Tool #4 (Introduce a Variable) keeps the bookkeeping honest by naming the midpoint and the doubled point so every length can be tracked.

1STEP 1

Ask what the domain really means

Note x is the median's length.

M = midpoint of BC, x = AM, f(x) = [△ ABC]

2STEP 2

Double the median

Extending makes a parallelogram.

MD = AM → ABDC is a parallelogram → BD = AC = 42, AD = 2x

3STEP 3

Read off the domain

The triangle inequality gives between 1 and 41.

|42 - 40| ＜ 2x ＜ 42 + 40 → 2 ＜ 2x ＜ 82 → 1 ＜ x ＜ 41

4STEP 4

Push the area to its maximum

Perpendicular sides give the peak 840.

[△ ABC] = 1/2 · 40 · h ≤ 1/2 · 40 · 42 = 840

5STEP 5

Find the median at that maximum

There the median is 29.

BC = √(40² + 42²) = √(3364) = 58, x = BC/2 = 29

6STEP 6

Add the four numbers

One plus 41 plus 840 plus 29 is 911.

p + q + r + s = 1 + 41 + 840 + 29 = 911 (C)

Answer

911

Both endpoints pass a sanity test against the two flattened pictures. If B and C collapse onto the same ray from A, the midpoint M sits (40 + 42)/2 = 41 away from A, matching q = 41; if they collapse onto opposite rays, M sits (42 - 40)/2 = 1 away, matching p = 1. A spot check inside the interval: x = 29 forces BC = 58, and the triple (40, 42, 58) satisfies Heron with semiperimeter 70, giving √(70 · 30 · 28 · 12) = √(705600) = 840 — exactly r. Just outside, x = 0.9 would force BC ≈ 82.004 ＞ 40 + 42 and x = 41.1 would force BC² ＜ 0, so neither is drawable, confirming the interval is open at both ends.

💡Key takeaway

Extend the median until it doubles: the picture becomes a triangle with sides 40, 42, and 2x, so the triangle inequality alone gives the domain (1, 41), and the area peaks at 840 when the right angle at A makes the median 29.

Ask what the domain really means
Double the median
Read off the domain
Push the area to its maximum
Find the median at that maximum
Add the four numbers