AMC 10 · 2020 · #8

Grade 8 geometry-2d

Archetype Casework by Position / Vertex Type

area-trianglesspatial-visualizationperpendicular-bisector caseworkidentify-subproblemssymmetry-argument ↑ Prerequisites: area-trianglespythagorean-theorem

📏 Long solution 💡 3 insights

Problem

Points $P$ and $Q$ lie in a plane with $PQ=8$ . How many locations for point $R$ in this plane are there such that the triangle with vertices $P$ , $Q$ , and $R$ is a right triangle with area $12$ square units?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

View mode:

Toolkit + CCSS Solution

Understand

Restated: $P$ and $Q$ are fixed points in the plane with $PQ = 8$. Count the locations of $R$ that make $\triangle PQR$ a right triangle of area $12$.

Givens: $PQ = 8$; $\triangle PQR$ must be a right triangle (one angle is exactly $90^\circ$); Area of $\triangle PQR$ is $12$; Choices: (A) $2$, (B) $4$, (C) $6$, (D) $8$, (E) $12$

Unknowns: The number of distinct points $R$ that satisfy both conditions

Understand

Restated: $P$ and $Q$ are fixed points in the plane with $PQ = 8$. Count the locations of $R$ that make $\triangle PQR$ a right triangle of area $12$.

Givens: $PQ = 8$; $\triangle PQR$ must be a right triangle (one angle is exactly $90^\circ$); Area of $\triangle PQR$ is $12$; Choices: (A) $2$, (B) $4$, (C) $6$, (D) $8$, (E) $12$

Plan

Primary tool: #1 Draw a Diagram

Secondary: #7 Identify Subproblems, #2 Make a Systematic List, #3 Eliminate Possibilities

Tool #1 puts $P$ and $Q$ on a horizontal line so we can see the perpendicular conditions. Tool #7 splits the count into three cases by where the right angle sits ($P$, $Q$, or $R$). Tool #2 enumerates the positions in each case without missing reflections. Tool #3 matches the total against the choices.

Execute — Answer: D

#1 Draw a Diagram 6.G.A.1 Step 1

Place $P$ and $Q$ on a horizontal line, $8$ apart.
The area of $\triangle PQR$ equals $\tfrac12 \cdot \text{base} \cdot \text{height} = \tfrac12 \cdot 8 \cdot h = 4h$.
Set $4h = 12$ so the height of $R$ above (or below) line $PQ$ must be $h = 3$.
So $R$ lies on one of two horizontal lines at distance $3$ from line $PQ$.

$$\tfrac{1}{2} \cdot 8 \cdot h = 12 \;\Rightarrow\; h = 3$$

💡 Draw $PQ$ horizontal; the area pins $R$ to a height of $3$.

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

Case A — right angle at $P$.
Then $PR \perp PQ$, so $PR$ is vertical.
Area $= \tfrac12 \cdot PQ \cdot PR = \tfrac12 \cdot 8 \cdot PR = 12$, so $PR = 3$.
Two such $R$: $3$ units above $P$ and $3$ units below $P$.

$$PR \perp PQ,\quad PR = 3 \;\Rightarrow\; 2 \text{ positions}$$

💡 Leg is the perpendicular segment from $P$; up or down.

#7 Identify Subproblems 6.G.A.1 Step 3

Case B — right angle at $Q$.
Symmetric to Case A: $QR \perp PQ$ with $QR = 3$.
Two more $R$: $3$ above $Q$ and $3$ below $Q$.

$$QR \perp PQ,\quad QR = 3 \;\Rightarrow\; 2 \text{ positions}$$

💡 Same as Case A but anchored at $Q$.

#7 Identify Subproblems 8.G.B.7 Step 4

Case C — right angle at $R$.
Then $PQ$ is the hypotenuse, so $R$ lies on the circle with diameter $PQ$ (a $90^\circ$ angle is inscribed in a semicircle).
This circle has center at the midpoint $M$ of $PQ$ and radius $4$.
$R$ must also sit at height $h = 3$ from line $PQ$.

$$R \text{ on circle},\; \text{center } M,\; \text{radius } 4,\; \text{height } |y_R| = 3$$

💡 Right angle at $R$ $\Leftrightarrow$ $R$ on the circle with $PQ$ as diameter.

#2 Make a Systematic List 8.G.B.7 Step 5

Find where the circle meets the lines $y = 3$ and $y = -3$.
Place $M = (0,0)$, so $P = (-4, 0)$ and $Q = (4, 0)$, and the circle is $x^2 + y^2 = 16$.
Set $y = 3$: $x^2 = 16 - 9 = 7$, so $x = \pm\sqrt{7}$.
Two points above.
By symmetry, the line $y = -3$ also gives two points below.
Total in Case C: $4$.

$$x^2 = 16 - 9 = 7 \;\Rightarrow\; x = \pm\sqrt{7};\;\; 2 \text{ above} + 2 \text{ below} = 4$$

💡 Two heights, two horizontal solutions each — $4$ intersection points.

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

Add the three cases.
None of the $8$ points coincide: Case A points have $x = -4$, Case B have $x = 4$, Case C have $x = \pm\sqrt{7} \approx \pm 2.65$ — all distinct.

$$2 + 2 + 4 = 8$$

💡 Disjoint cases sum directly.

#3 Eliminate Possibilities 4.NBT.A.2 Step 7

$8$ matches choice (D).

$$8 \Rightarrow \textbf{(D)}$$

💡 Read the matching answer choice.

[1] #1 6.G.A.1 Place $P$ and $Q$ on a horizontal line, $8$ apart. The area of $\triangle PQR$ e

[2] #7 6.G.A.1 Case A — right angle at $P$. Then $PR \perp PQ$, so $PR$ is vertical. Area $= \t

[3] #7 6.G.A.1 Case B — right angle at $Q$. Symmetric to Case A: $QR \perp PQ$ with $QR = 3$. T

[4] #7 8.G.B.7 Case C — right angle at $R$. Then $PQ$ is the hypotenuse, so $R$ lies on the cir

[5] #2 8.G.B.7 Find where the circle meets the lines $y = 3$ and $y = -3$. Place $M = (0,0)$, s

[6] #7 2.OA.A.1 Add the three cases. None of the $8$ points coincide: Case A points have $x = -4

[7] #3 4.NBT.A.2 $8$ matches choice (D).

Review

Reasonableness: Symmetry check: the configuration is symmetric across line $PQ$ (so the total must be even) and across the perpendicular bisector of $PQ$ (the line $x = 0$). The eight points $(-4, \pm 3), (4, \pm 3), (\pm\sqrt{7}, \pm 3)$ are closed under both reflections. ✓ Each point gives a triangle: Case A — legs $8, 3$, area $= \tfrac12 \cdot 8 \cdot 3 = 12$ ✓; Case C — legs computed from Pythagorean theorem, $PR^2 + QR^2 = 64$ with $PR \cdot QR = 24$ (since area $= 12$), consistent. ✓

Alternative: Tool #13 (Algebra): place $P = (0,0)$ and $Q = (8, 0)$. Let $R = (x, y)$ with $|y| = 3$. For a right angle at $R$: $\vec{RP} \cdot \vec{RQ} = 0$ gives $x(x - 8) + y^2 = 0$, i.e. $x^2 - 8x + 9 = 0$, so $x = 4 \pm \sqrt{7}$ — two $x$-values per choice of sign of $y$, giving $4$ Case-C points (same count via coordinates instead of circle geometry).

CCSS standards used (min grade 8)

6.G.A.1 Find area of triangles, special quadrilaterals, and polygons by composing (Using $\text{area} = \tfrac12 \cdot \text{base} \cdot \text{height}$ to pin the height of $R$ at $3$ and to handle the right-angle-at-$P$ and right-angle-at-$Q$ cases.)
8.G.B.7 Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Case C: identifying that a right angle at $R$ forces $R$ on the circle of diameter $PQ$ (via the Pythagorean relation $PR^2 + QR^2 = PQ^2$) and finding the four intersections with $y = \pm 3$.)
2.OA.A.1 Solve one- and two-step word problems using addition and subtraction within 100 (Adding the three case counts: $2 + 2 + 4 = 8$.)
4.NBT.A.2 Read and write multi-digit whole numbers and compare using symbols (Matching the final count $8$ to choice (D).)

⭐ This AMC 10 problem only needs Grade 8 right-triangle thinking you already know — the area $12$ forces $R$ to be $3$ units away from line $PQ$. Then the right angle can sit at $P$ ($2$ spots), at $Q$ ($2$ spots), or at $R$ on the circle with $PQ$ as diameter ($4$ spots). Total $8$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 8)

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