AMC 10 · 2018 · #23

Grade 10 geometry-2d

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

triangle-midsegmentrotation-isometryisosceles-triangleangle-sum-trianglemidpoint-formula spatial-visualizationidentify-subproblems ↑ Prerequisites: angle-sum-triangleisosceles-trianglemidpoint-formula

📏 Long solution 💡 4 insights

Problem

Triangle PAT has angle P of 36 degrees, angle A of 56 degrees, and base PA of 10. On side TP mark U with PU of 1, and on side TA mark G with AG of 1, so U sits just above P and G just above A. Let M be the midpoint of PA and N the midpoint of UG. Find the acute angle between line MN and line PA.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

How to solve

Strategy Visualize Spatial Relationships

Segment MN is short, tilted, and has no angle attached to it, so there is nothing to chase directly. Tool #17 (Visualize Spatial Relationships) supplies the key move: spin the little corner triangle PUM a half turn about M. That half turn drags the length PU=1 over next to the length AG=1, and suddenly two equal sides meet at a single vertex. Tool #1 (Draw a Diagram) keeps track of which side of line PA each new point lands on. Tool #7 (Identify Subproblems) splits the work into two small jobs: show MN is a midsegment, then find the tilt of the line it is parallel to. Tool #16 (Change Focus) is the payoff: measure the angle on that easier parallel line instead of on MN.

1STEP 1

Draw it and fill in angle T

Fill in the third angle first.

∠ T = 180°-36°-56° = 88°

2STEP 2

Spin the corner a half turn about M

Spin a half turn about the midpoint.

AU' = PU = 1, and M is the midpoint of UU'

3STEP 3

MN turns into a midsegment

The two midpoints form a midsegment.

M,N midpoints in △ UU'G → MN ∥ U'G

4STEP 4

The angle at A is 92 degrees

After the spin one angle becomes the sum of two.

∠ U'AG = ∠ U'AP + ∠ PAG = 36° + 56° = 92°

5STEP 5

Isosceles gives base angles of 44

An isosceles triangle gives the base angles.

∠ AGU' = ∠ AU'G = (180°-92°)/2 = 44°

6STEP 6

Where U'G crosses the base

Read the angle where it meets the base.

∠ GHA = 180° - 56° - 44° = 80°

7STEP 7

Carry the angle back to MN

By parallelism the answer is 80 degrees.

MN ∥ U'G → acute angle between MN and PA = 80°

Answer

Neither PA=10 nor PU=AG=1 survives into the final answer, which is the tell that only the equality PU=AG ever mattered. Chasing that hint, MN turns out to be parallel to the bisector of ∠ T: the bisector splits 88° into two 44° halves, and the triangle it forms with A has angles 56° and 44°, leaving 80° again. A direct coordinate check agrees: with P=(0,0) and A=(10,0), the slope of MN is (sin 36°+sin 56°)/(cos 36°-cos 56°)≈1.4168/0.2498≈ 5.671, and tan 80°≈ 5.671. A slope that steep also matches the drawing, where MN climbs sharply. One trap worth naming: several routes naturally produce ∠ NMP=100°, but the problem asks for the acute angle, so the supplement 80° is what gets reported.

💡Key takeaway

When a segment has no angles attached to it, spin part of the picture a half turn until the equal lengths meet, then measure a parallel line instead.

Draw it and fill in angle T
Spin the corner a half turn about M
MN turns into a midsegment
The angle at A is 92 degrees
Isosceles gives base angles of 44
Where U'G crosses the base
Carry the angle back to MN