AMC 10 · 2025 · #20

Grade 8 geometry-2d

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

coordinate-geometryquadratic-equationsslope-intercept convert-to-algebra ↑ Prerequisites: coordinate-geometryquadratic-equations

📏 Long solution 💡 3 insights

Problem

A round silo of diameter 20 m (radius 10 m) sits in a field. From the silo's center O, MacDonald stands 20 m west and 15 m south, and McGregor stands 20 m east and g ＞ 0 m south. The straight line of sight between the two people just grazes (is tangent to) the silo. Write g=(a√(b)-c)/d in lowest admissible form and report a+b+c+d.

Pick an answer.

(A)

119

(B)

120

(C)

121

(D)

122

(E)

123

AMC 10 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Convert to Algebra

The words 'west', 'south', 'center', and 'tangent' all describe positions and a circle, which is exactly what a coordinate grid handles. Tool #13 (Convert to Algebra) turns the field into an xy-plane: the silo becomes a circle equation and the line of sight becomes a line. The one fact still missing is the direction of the line of sight, so Tool #4 (Introduce a Variable) names its slope m and lets the tangency condition solve for it. Tangency has a clean algebraic form: the distance from the center to the line equals the radius. That distance equation squares out to a quadratic in m with two roots, and Tool #3 (Eliminate Possibilities) uses the requirement g ＞ 0 to throw out the wrong one. Tool #1 (Draw a Diagram) keeps the compass directions straight so north/south signs are correct.

1STEP 1

Put the field on a grid

Put MacDonald at the origin with east as +x: then O=(20,15) and McGregor sits at (40,15-g).

O=(20,15), silo: (x-20)²+(y-15)²=100, McGregor=(40,15-g)

2STEP 2

Name the line of sight's slope

The sight line goes through the origin, so it is y=mx with m=(15-g)/40; leave m unknown for now.

y=mx, m=(15-g)/40

3STEP 3

Turn tangency into a distance equation

Tangency says O sits one radius from the line mx-y=0, so (|20m-15|)/(√(m²+1))=10.

(|20m-15|)/(√(m²+1))=10

4STEP 4

Square and simplify to a quadratic

Squaring kills the root and the absolute value; the m² terms partly cancel, leaving 12m²-24m+5=0.

(20m-15)²=100(m²+1) → 400m²-600m+225=100m²+100 → 300m²-600m+125=0 → 12m²-24m+5=0

5STEP 5

Solve for the slope

The discriminant is 576-240=336=16·21, so √(336)=4√(21) and m=(6±√(21))/6.

m=24±√(336)/24=24±4√(21)/24=6±√(21)/6

6STEP 6

Pick the root that keeps g positive

Since g=15-40m, the slope near 1.76 drives g negative; only m=(6-√(21))/6≈0.236 survives.

m=(6+√(21))/6→ g ＜ 0 (rejected), m=(6-√(21))/6→ g ＞ 0 (kept)

7STEP 7

Recover g and add the pieces

Substituting gives g=(20√(21)-75)/3, so a=20, b=21, c=75, d=3 and the sum is 119.

g=15-(120-20√(21))/3=(45-120+20√(21))/3=(20√(21)-75)/3 → a+b+c+d=119 (A)

Answer

119

Numerically g=(20√(21)-75)/3≈(20(4.583)-75)/3=(91.65-75)/3≈5.55 ＞ 0, which respects g ＞ 0. Back-check the slope: m=(15-5.55)/40=9.45/40≈0.236=(6-√(21))/6, matching Step 6. The distance from O=(20,15) to the line y≈0.236x is (|20(0.236)-15|)/(√(0.236²+1))=10.28/1.028≈10, equal to the radius, confirming tangency. The admissible-form checks (b squarefree, d coprime to gcd(a,c)) all hold, and the total 119 is choice (A).

💡Key takeaway

Drop the picture onto a grid: a line that just grazes a circle sits exactly one radius from the center, and that single fact turns 'tangent' into an equation you can solve.

Put the field on a grid
Name the line of sight's slope
Turn tangency into a distance equation
Square and simplify to a quadratic
Solve for the slope
Pick the root that keeps g positive
Recover g and add the pieces