AMC 10 · 2025 · #20
Grade 8 geometry-2dA silo (right circular cylinder) with diameter 20 meters stands in a field. MacDonald is located 20 meters west and 15 meters south of the center of the silo. McGregor is located 20 meters east and g>0 meters south of the center of the silo. The line of sight between MacDonald and McGregor is tangent to the silo. The value of g can be written as dab−c, where a,b,c, and d are positive integers, b is not divisible by the square of any prime, and d is relatively prime to the greatest common divisor of a and c. What is a+b+c+d?
Pick an answer.
AMC 10 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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 $g=\frac{a\sqrt{b}-c}{d}$ in lowest admissible form and report $a+b+c+d$.
Givens: The silo is a circle of radius $10$ m centered at $O$; MacDonald is $20$ m west and $15$ m south of $O$; McGregor is $20$ m east and $g>0$ m south of $O$; The line joining MacDonald and McGregor is tangent to the silo; $g=\frac{a\sqrt{b}-c}{d}$ with $a,b,c,d$ positive integers, $b$ squarefree, and $d$ coprime to $\gcd(a,c)$; Answer choices: (A) $119$, (B) $120$, (C) $121$, (D) $122$, (E) $123$
Unknowns: The value of $g$, and from it the sum $a+b+c+d$
Understand
Restated: 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 $g=\frac{a\sqrt{b}-c}{d}$ in lowest admissible form and report $a+b+c+d$.
Givens: The silo is a circle of radius $10$ m centered at $O$; MacDonald is $20$ m west and $15$ m south of $O$; McGregor is $20$ m east and $g>0$ m south of $O$; The line joining MacDonald and McGregor is tangent to the silo; $g=\frac{a\sqrt{b}-c}{d}$ with $a,b,c,d$ positive integers, $b$ squarefree, and $d$ coprime to $\gcd(a,c)$; Answer choices: (A) $119$, (B) $120$, (C) $121$, (D) $122$, (E) $123$
Plan
Primary tool: #13 Convert to Algebra
Secondary: #4 Introduce a Variable, #3 Eliminate Possibilities, #1 Draw a Diagram
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.
Execute — Answer: A
8.G.B.8 Step 1 Put the field on a grid
- Set MacDonald at the origin and let east be $+x$, north be $+y$.
- MacDonald is $20$ west and $15$ south of the center, so the center is $20$ east and $15$ north of MacDonald: $O=(20,15)$.
- McGregor is $20$ east and $g$ south of $O$, so McGregor $=(40,\,15-g)$.
- The silo is the circle of radius $10$ about $O$.
💡 A compass description becomes exact once you nail down which way is $+x$ and $+y$.
8.EE.B.6 Step 2 Name the line of sight's slope
- The line of sight passes through MacDonald at the origin, so it has the form $y=mx$ for some slope $m$.
- Because it must also reach McGregor $(40,15-g)$, the slope is tied to $g$ by rise over run.
- We keep $m$ as the unknown for now; the tangency condition will pin it down.
💡 A line through the origin is captured by a single number, its slope.
8.G.B.8 Step 3 Turn tangency into a distance equation
- A line is tangent to a circle exactly when the perpendicular distance from the center to the line equals the radius.
- Write the line $y=mx$ as $mx-y=0$.
- The distance from $O=(20,15)$ to this line is $\frac{|20m-15|}{\sqrt{m^2+1}}$, and tangency forces it to equal the radius $10$.
💡 Tangent means 'just touching', so the center sits exactly one radius away from the line.
8.EE.C.7 Step 4 Square and simplify to a quadratic
- Squaring both sides clears the square root and the absolute value.
- Expand and collect all terms on one side; the $m^2$ terms partly cancel, leaving a clean quadratic in $m$ after dividing through by the common factor.
💡 Squaring trades an awkward root-and-absolute-value equation for an ordinary polynomial.
8.EE.A.2 Step 5 Solve for the slope
- Apply the quadratic formula to $12m^2-24m+5=0$.
- The number under the root is $24^2-4\cdot12\cdot5=576-240=336=16\cdot21$, so $\sqrt{336}=4\sqrt{21}$, and the $4$ cancels the common factor with $24$ and $24$.
💡 Two possible slopes appear because two lines through MacDonald can graze the silo.
8.NS.A.2 Step 6 Pick the root that keeps g positive
- From Step 2, $g=15-40m$.
- Since $\sqrt{21}\approx4.58$, the two slopes are about $m\approx1.76$ and $m\approx0.236$.
- The larger slope gives $g\approx15-40(1.76)<0$, which breaks the rule $g>0$, so it is rejected.
- The smaller slope $m=\frac{6-\sqrt{21}}{6}$ gives a positive $g$ and is the one we want.
💡 Estimating $\sqrt{21}$ as a decimal instantly shows which line points the wrong way.
8.EE.C.7 Step 7 Recover g and add the pieces
- Substitute the kept slope into $g=15-40m$.
- Then $40m=\frac{40(6-\sqrt{21})}{6}=\frac{120-20\sqrt{21}}{3}$, and subtracting from $15=\frac{45}{3}$ gives $g=\frac{20\sqrt{21}-75}{3}$.
- Matching $\frac{a\sqrt{b}-c}{d}$ reads off $a=20$, $b=21$, $c=75$, $d=3$.
- Check: $b=21=3\cdot7$ is squarefree, $\gcd(20,75)=5$, and $\gcd(d,5)=\gcd(3,5)=1$, so the form is admissible.
- The sum is $20+21+75+3=119$, choice $\textbf{(A)}$.
💡 Once the slope is fixed, $g$ is just plug-and-simplify.
8.G.B.8 Set MacDonald at the origin and let east be $+x$, north be $+y$. MacDonald is $2 8.EE.B.6 The line of sight passes through MacDonald at the origin, so it has the form $y= 8.G.B.8 A line is tangent to a circle exactly when the perpendicular distance from the c 8.EE.C.7 Squaring both sides clears the square root and the absolute value. Expand and co 8.EE.A.2 Apply the quadratic formula to $12m^2-24m+5=0$. The number under the root is $24 8.NS.A.2 From Step 2, $g=15-40m$. Since $\sqrt{21}\approx4.58$, the two slopes are about 8.EE.C.7 Substitute the kept slope into $g=15-40m$. Then $40m=\frac{40(6-\sqrt{21})}{6}=\ Review
Reasonableness: Numerically $g=\frac{20\sqrt{21}-75}{3}\approx\frac{20(4.583)-75}{3}=\frac{91.65-75}{3}\approx5.55>0$, which respects $g>0$. Back-check the slope: $m=\frac{15-5.55}{40}=\frac{9.45}{40}\approx0.236=\frac{6-\sqrt{21}}{6}$, matching Step 6. The distance from $O=(20,15)$ to the line $y\approx0.236x$ is $\frac{|20(0.236)-15|}{\sqrt{0.236^2+1}}=\frac{10.28}{1.028}\approx10$, 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).
Alternative: Instead of the distance formula, substitute $y=mx$ directly into the circle equation to get $(m^2+1)x^2-(30m+40)x+525=0$, then set the discriminant to zero (a tangent line meets the circle in one point). That discriminant condition simplifies to the same $12m^2-24m+5=0$, so the rest of the work — and the answer $119$ — is identical. This route trades the memorized point-to-line distance formula for the single-root idea, useful if you don't recall the distance formula.
CCSS standards used (min grade 8)
8.G.B.8Apply the Pythagorean theorem to find distance between two points in a coordinate system (Placing the field on a coordinate grid and writing the point-to-line distance from the silo's center to the line of sight.)8.EE.B.6Use similar triangles to explain why the slope is the same between any two points (Representing the line of sight through the origin as $y=mx$ and tying the slope $m$ to $g$ via rise over run.)8.EE.C.7Solve linear equations in one variable (Clearing the distance equation to $12m^2-24m+5=0$ and then solving $g=15-40m$ for $g$.)8.EE.A.2Use square root and cube root symbols to represent solutions (Simplifying $\sqrt{336}=4\sqrt{21}$ in the quadratic formula to express the slope $m=\frac{6\pm\sqrt{21}}{6}$.)8.NS.A.2Use rational approximations of irrational numbers to compare their size (Approximating $\sqrt{21}\approx4.58$ to decide which slope keeps $g>0$ and to sanity-check the final value.)
⭐ 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.
⭐ 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.
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