AMC 10 · 2008 · #17

Grade 10 geometry-2d

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

coordinate-geometryperpendicular-slopesdifference-of-squaresdigit-sum convert-to-algebracasework ↑ Prerequisites: coordinate-geometryarea-triangles

📏 Long solution 💡 3 insights

Problem

Three points on a parabola form a right triangle with one horizontal side and a known area. Find the digit sum of the third point's height.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

How to solve

Strategy Convert to Algebra

Every point on the curve is (t, t²), so one letter per point captures everything. The parabola is symmetric about the y-axis, which pins A and B to be mirror images and gives the base for free. The phrase 'right triangle' does not say where the right angle is, so the first real job is to rule out two of the three vertices; that is a logic step, not a computation. Once the right angle is located at C, writing perpendicularity as an algebraic equation turns the whole problem into one factorable expression, and the surprise it reveals makes the area condition collapse to a single line.

1STEP 1

Name each point with one letter

The horizontal side makes the two base points symmetric.

A=(-m, m²), B=(m, m²) (m ＞ 0), C=(n, n²), n² ≠ m²

2STEP 2

The right angle can only be at C

The right angle cannot sit at a base vertex.

∠ A = 90° → AC ⊥ AB → AC vertical → x_C = x_A → C=A (contradiction)

3STEP 3

Write the right angle at C as an equation

Writing it at the apex gives one equation.

CA=(-m-n, m²-n²), CB=(m-n, m²-n²), CA·CB=0

4STEP 4

Factor: the height is forced to be 1

Factoring forces the triangle's height to be exactly one.

-(m²-n²)+(m²-n²)²=0 → (m²-n²)[(m²-n²)-1]=0 → m²-n²=1

5STEP 5

Area collapses to a single letter

The area then collapses to a single letter.

AB = 2m, h = |m²-n²| = 1, [ABC]=1/2(2m)(1)=m=2008

6STEP 6

Check the triangle actually exists

Such a triangle really exists.

n=±√(2008²-1), [ABC]=1/2 · 4016 · 1 = 2008 ✓

7STEP 7

Compute the y-coordinate and add its digits

Adding the digits gives 18, choice (A).

y_C=n²=2008²-1=(2000+8)²-1=4032064-1=4032063, 4+0+3+2+0+6+3=18

Answer

The shape is extreme but legal: base 4016 and height 1, a sliver of a triangle whose area is half of 4016, namely 2008. That matches the required area, and the right angle at C is real because the two legs from C have slopes n - m and n + m whose product is n² - m² = -1. The digit sum 18 lands inside the offered range of 16 to 20, and the neighboring choices are exactly the traps: forgetting the minus one and using 2008² = 4032064 gives digit sum 19, which is choice (D). Using m itself, 2008, gives 10, which is not offered, a sign that the intended object is n² and not m. Dropping the one-half in the area formula would give m = 1004 and digit sum 15, also not offered. The construction in step 6 shows the configuration truly exists, so 18 is achieved and not merely forced.

💡Key takeaway

On y = x² the right angle has nowhere to go but C, and that single fact locks the triangle's height at exactly 1, so the area is just half the base.

Name each point with one letter
The right angle can only be at C
Write the right angle at C as an equation
Factor: the height is forced to be 1
Area collapses to a single letter
Check the triangle actually exists
Compute the y-coordinate and add its digits