AMC 10 · 2019 · #10

Grade 8 geometry-2d

Archetype Small-Domain Constrained Search

pythagorean-theoremcoordinate-geometryarea-trianglesperimeterpolygon-inequality identify-subproblemscasework ↑ Prerequisites: pythagorean-theoremarea-trianglescoordinate-geometry

📏 Medium solution 💡 3 insights

Problem

In a given plane, points $A$ and $B$ are $10$ units apart. How many points $C$ are there in the plane such that the perimeter of $\triangle ABC$ is $50$ units and the area of $\triangle ABC$ is $100$ square units?

$\textbf{(A) }0\qquad\textbf{(B) }2\qquad\textbf{(C) }4\qquad\textbf{(D) }8\qquad\textbf{(E) }\text{infinitely many}$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

infinitely many

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Toolkit + CCSS Solution

Understand

Restated: Points $A$ and $B$ are $10$ units apart in a plane. Count all points $C$ such that $\triangle ABC$ has perimeter exactly $50$ and area exactly $100$.

Givens: $|AB| = 10$; Perimeter of $\triangle ABC = 50$; Area of $\triangle ABC = 100$; Choices: (A) $0$, (B) $2$, (C) $4$, (D) $8$, (E) $\text{infinitely many}$

Unknowns: The number of points $C$ satisfying both conditions

Understand

Restated: Points $A$ and $B$ are $10$ units apart in a plane. Count all points $C$ such that $\triangle ABC$ has perimeter exactly $50$ and area exactly $100$.

Givens: $|AB| = 10$; Perimeter of $\triangle ABC = 50$; Area of $\triangle ABC = 100$; Choices: (A) $0$, (B) $2$, (C) $4$, (D) $8$, (E) $\text{infinitely many}$

Plan

Primary tool: #1 Draw a Diagram

Secondary: #9 Solve an Easier Related Problem, #7 Identify Subproblems, #6 Guess and Check, #3 Eliminate Possibilities

Plant $AB$ on a coordinate axis (Tool #1 + #9 makes the picture concrete). Split the requirements into (a) area $\Rightarrow$ height of $C$ above $AB$, and (b) perimeter $\Rightarrow$ sum of slanted sides $AC + BC$ (Tool #7 sub-questions). Then test whether ANY $C$ at the required height can give the required slant sum — try the best (closest) candidate first (Tool #6); if even that fails, no $C$ works.

Execute — Answer: A

#1 Draw a Diagram 8.G.B.8 Step 1

Place coordinates: $A = (-5, 0)$, $B = (5, 0)$, so $|AB| = 10$ lies on the $x$-axis.
Let $C = (x, y)$.

$$A = (-5, 0),\; B = (5, 0),\; C = (x, y)$$

💡 Setting $AB$ on the $x$-axis turns geometry into easy distance calculations.

#7 Identify Subproblems 6.G.A.1 Step 2

Sub-question 1 (area).
Take $AB$ as the base of length $10$.
The triangle's height above $AB$ is $|y|$, so area $= \tfrac{1}{2} \cdot 10 \cdot |y| = 5|y|$.
Setting area $= 100$ gives $|y| = 20$.
So $C$ must sit on one of the two horizontal lines $y = 20$ or $y = -20$.

$$5|y| = 100 \;\Rightarrow\; |y| = 20$$

💡 Area = $\tfrac{1}{2}\cdot \text{base} \cdot \text{height}$ pins down how far $C$ is from $AB$.

#7 Identify Subproblems 6.EE.B.7 Step 3

Sub-question 2 (perimeter).
Perimeter $= |AB| + |AC| + |BC| = 10 + |AC| + |BC| = 50$.
So we need $|AC| + |BC| = 40$.

$$|AC| + |BC| = 50 - 10 = 40$$

💡 The two slanted sides must add to $40$.

#6 Guess and Check 8.G.B.8 Step 4

Test the easiest candidate: $C = (0, 20)$ (directly above the midpoint of $AB$, the symmetric choice).
By the Pythagorean theorem, $|AC| = |BC| = \sqrt{5^2 + 20^2} = \sqrt{425} = 5\sqrt{17}$.
Sum $= 10\sqrt{17}$.

$$|AC| + |BC| = 10\sqrt{17}$$

💡 The symmetric point above the base is usually the trickiest — easiest place to test the perimeter.

#6 Guess and Check 8.NS.A.2 Step 5

Estimate: $\sqrt{17} \approx 4.123$, so $10\sqrt{17} \approx 41.23$.
That is already MORE than $40$.

$$10\sqrt{17} \approx 41.23 > 40$$

💡 Even the smallest possible $|AC|+|BC|$ overshoots the budget.

#9 Solve an Easier Related Problem 8.G.B.7 Step 6

Sub-question 3: is $C = (0, 20)$ actually the MINIMUM of $|AC| + |BC|$ along the line $y = 20$?
Yes — reflect $A$ across $y = 20$ to $A' = (-5, 40)$.
Then $|AC| + |BC| = |A'C| + |BC| \ge |A'B|$ (triangle inequality), with equality when $A', C, B$ are collinear.
Here $|A'B| = \sqrt{10^2 + 40^2} = \sqrt{1700} = 10\sqrt{17}$ — same value, hit only at the symmetric point.
So $|AC| + |BC| \ge 10\sqrt{17} > 40$ for every $C$ on $y = 20$.

$$\min_{C \text{ on } y=20}(|AC|+|BC|) = 10\sqrt{17} > 40$$

💡 Going further left or right only stretches the slanted distances — symmetric point is the tightest.

#3 Eliminate Possibilities K.MD.B.3 Step 7

By symmetry the same holds on $y = -20$.
So NO point $C$ at all can satisfy both perimeter $= 50$ AND area $= 100$.
The answer is $0$.

$$\text{Count of valid } C = 0 \;\Rightarrow\; \textbf{(A)}$$

💡 If even the best candidate fails, no point works — answer is $0$.

[1] #1 8.G.B.8 Place coordinates: $A = (-5, 0)$, $B = (5, 0)$, so $|AB| = 10$ lies on the $x$-a

[2] #7 6.G.A.1 Sub-question 1 (area). Take $AB$ as the base of length $10$. The triangle's heig

[3] #7 6.EE.B.7 Sub-question 2 (perimeter). Perimeter $= |AB| + |AC| + |BC| = 10 + |AC| + |BC| =

[4] #6 8.G.B.8 Test the easiest candidate: $C = (0, 20)$ (directly above the midpoint of $AB$,

[5] #6 8.NS.A.2 Estimate: $\sqrt{17} \approx 4.123$, so $10\sqrt{17} \approx 41.23$. That is alr

[6] #9 8.G.B.7 Sub-question 3: is $C = (0, 20)$ actually the MINIMUM of $|AC| + |BC|$ along the

[7] #3 K.MD.B.3 By symmetry the same holds on $y = -20$. So NO point $C$ at all can satisfy both

Review

Reasonableness: Quick gut check: a triangle with base $10$ and area $100$ has height $20$ — it's a very TALL skinny triangle. The shortest slant from a point $20$ up to either endpoint of a base $10$ wide is already at least $\sqrt{5^2 + 20^2} \approx 20.6$. Two of those slants sum to at least $\approx 41.2$, plus the base $10$ gives perimeter at least $\approx 51.2 > 50$. There's no room — answer (A) $0$ is consistent.

Alternative: Tool #13 (Algebra) via the ellipse: "perimeter $=50$" means $|AC|+|BC|=40$, the locus of $C$ is an ellipse with foci $A, B$ and major axis $40$. Its semi-major is $20$ and (since foci are $\pm 5$) semi-minor is $\sqrt{20^2-5^2} = \sqrt{375} \approx 19.36$. The maximum height of any $C$ on this ellipse is $\approx 19.36 < 20$, so the area-$100$ requirement ($|y|=20$) can never be met. Same answer: $0$.

CCSS standards used (min grade 8)

8.G.B.8 Apply the Pythagorean theorem to find distance between two points in a coordinate system (Computing $|AC|$ and $|BC|$ from coordinates of $A$, $B$, and $C$.)
8.G.B.7 Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Using triangle inequality with the reflected point $A'$ to show $|AC| + |BC| \ge 10\sqrt{17}$ along the line $y = 20$.)
8.NS.A.2 Use rational approximations of irrational numbers to compare their size (Approximating $10\sqrt{17} \approx 41.23$ and comparing to $40$.)
6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing (Using area $= \tfrac{1}{2}\cdot\text{base}\cdot\text{height}$ to translate area $= 100$ into $|y| = 20$.)
6.EE.B.7 Solve real-world problems by writing and solving equations of the form px = q (Solving the small linear equation $5|y| = 100$ for the height.)
K.MD.B.3 Classify objects into given categories and count the numbers in each (Counting how many candidate points pass both tests — which here is $0$.)

⭐ This AMC 10 problem only needs Grade 8 Pythagorean thinking you already know: area $=100$ forces $C$ to sit $20$ above (or below) $AB$. Even the closest such $C$ gives slants summing to $10\sqrt{17}\approx 41.2 > 40$, so the perimeter $50$ is impossible — $0$ points.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: A

Review

CCSS standards used (min grade 8)

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