AMC 10 · 2021 · #11

Grade 8 geometry-2d

Archetype Coordinate Plane Figure Computation └ Sub-archetype Line Intersection Point as the Answer

coordinate-geometryslope-interceptpythagorean-theoremcasework convert-to-algebraidentify-subproblemscasework ↑ Prerequisites: coordinate-geometryslope-intercept

📏 Long solution 💡 3 insights

Problem

A triangle has sides 13, 14, and 15. A point on one side splits it into pieces of length 5 and 10. On the line joining that point to the opposite vertex sit two points, each making a four-vertex figure a trapezoid. Exactly two such points exist. Find the distance between them.

Pick an answer.

(A)

$\frac{42}5$

(B)

$6\sqrt2$

(C)

$\frac{84}5$

(D)

$12\sqrt2$

(E)

How to solve

Strategy Draw a Diagram

Tool #2 (Systematic List) does the thinking first: a quadrilateral ABCD has only two pairs of opposite sides, {AB, CD} and {BC, AD}, so "trapezoid" can happen in exactly two ways. That is where the word "exactly two points" in the problem comes from, and it splits the work into two clean cases. Tool #1 (Draw a Diagram) then turns the picture into a coordinate grid — the 13-14-15 triangle has rational coordinates, so every point in this problem lands on a nice fraction. After that, Tool #13 (Convert to Algebra) makes each case a one-line question: intersect line BP with a line of known slope through a known point. Tool #7 (Identify Subproblems) orders the work: coordinates of B, then line BP, then D, then E, then one distance.

1STEP 1

Split into two cases

There are two ways to be a trapezoid.

Case 1: CD ∥ AB Case 2: AD ∥ BC

2STEP 2

Set up coordinates

The three sides fix the coordinates.

x² + y² &= 169 ; (x-15)² + y² &= 196 ; → -30x + 225 &= 27 ; x &= 33/5, y = √(169 - 1089/25) = 56/5

3STEP 3

Find the shared line

Find the line both points sit on.

P = (5,0), m_BP = (56/5 - 0)/(33/5 - 5) = 56/5/8/5 = 7 → y = 7(x-5)

4STEP 4

Find the first point

The first parallel condition gives a crossing.

m_AB &= 56/33, y = 56/33(x-15) ; 7(x-5) &= 56/33(x-15) ; 231(x-5) &= 56(x-15) ; 175x &= 315 → x = 9/5, y = 7(9/5-5) = -112/5

5STEP 5

Find the second point

The second works the same way.

m_BC &= (0 - 56/5)/(15 - 33/5) = -56/5/42/5 = -4/3, y = -4/3x ; 7(x-5) &= -4/3x ; 21x - 105 &= -4x → x = 21/5, y = -28/5

6STEP 6

Measure between them

The distance is twelve root two.

DE &= √((21/5-9/5)² + (-28/5+112/5)²) = √((12/5)² + (84/5)²) ; &= 12/5√(1 + 49) = 12/5 · 5√(2) = 12√(2) → (D)

Answer

12√2

Three checks. (1) Both points must sit on line BP: the gaps give slope (84/5)/(12/5) = 7, exactly m_BP, so D, E, B, P really are collinear. (2) The parallel conditions hold: slope of CD is (0 + 112/5)/(15 - 9/5) = 112/5/66/5 = 56/33 = m_AB, and slope of AE is -28/5/21/5 = -4/3 = m_BC. (3) Neither figure secretly became a parallelogram (which would have two parallel pairs and break the "exactly two points" count): slope of AD is -112/9 ≠ -4/3, and slope of CE is 28/54 = 14/27 ≠ 56/33. Also note the trap answers: 84/5 is only the vertical gap and 42/5 is half of it — both are pieces of the calculation, not the distance.

💡Key takeaway

This AMC 12 problem runs on Grade 8 slope-and-distance work. A quadrilateral has only two pairs of opposite sides, so "make ABCD a trapezoid" has only two ways to happen; drop the 13-14-15 triangle onto a grid, cross line BP with each parallel line, and the two hits are 12/5 apart sideways and 84/5 apart up-and-down, giving DE = 12√(2).