AMC 10 · 2021 · #24
Grade 8 geometry-2dProblem
The interior of a quadrilateral is bounded by the graphs of and , where is a positive real number. What is the area of this region in terms of , valid for all ?
Pick an answer.
Toolkit + CCSS Solution
Understand
Restated: The region enclosed by the graphs of $(x + a y)^2 = 4 a^2$ and $(a x - y)^2 = a^2$ (with $a > 0$) is a quadrilateral. Find its area as a function of $a$, valid for all $a > 0$.
Givens: $(x + a y)^2 = 4 a^2$, which factors into two parallel lines $x + a y = 2a$ and $x + a y = -2a$; $(a x - y)^2 = a^2$, which factors into two parallel lines $a x - y = a$ and $a x - y = -a$; $a > 0$; Answer choices: (A) $\tfrac{8a^2}{(a+1)^2}$, (B) $\tfrac{4a}{a+1}$, (C) $\tfrac{8a}{a+1}$, (D) $\tfrac{8a^2}{a^2+1}$, (E) $\tfrac{8a}{a^2+1}$
Unknowns: Area of the quadrilateral as a function of $a$
Understand
Restated: The region enclosed by the graphs of $(x + a y)^2 = 4 a^2$ and $(a x - y)^2 = a^2$ (with $a > 0$) is a quadrilateral. Find its area as a function of $a$, valid for all $a > 0$.
Givens: $(x + a y)^2 = 4 a^2$, which factors into two parallel lines $x + a y = 2a$ and $x + a y = -2a$; $(a x - y)^2 = a^2$, which factors into two parallel lines $a x - y = a$ and $a x - y = -a$; $a > 0$; Answer choices: (A) $\tfrac{8a^2}{(a+1)^2}$, (B) $\tfrac{4a}{a+1}$, (C) $\tfrac{8a}{a+1}$, (D) $\tfrac{8a^2}{a^2+1}$, (E) $\tfrac{8a}{a^2+1}$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems, #9 Solve an Easier Related Problem, #3 Eliminate Possibilities, #13 Convert to Algebra
Tool #1 (Diagram) — sketch the four lines; once you see two parallel pairs with perpendicular slopes, the region is obviously a rectangle. Tool #7 (Subproblems) — separately compute the rectangle's length (distance between $x + a y = \pm 2a$) and width (distance between $a x - y = \pm a$), then multiply. Tool #9 (Easier Problem) — plug $a = 1$ to get a concrete square, verify the answer-choice candidates collapse to the same number to spot-check the formula. Tool #3 (Eliminate) — $a = 1$ instantly knocks out wrong choices. Tool #13 (Algebra) — the distance-between-parallel-lines formula $\tfrac{|c_1 - c_2|}{\sqrt{A^2 + B^2}}$ for $A x + B y = c$ gives the cleanest derivation.
Execute — Answer: D
8.EE.A.2 Step 1 - Factor each equation into two parallel lines.
- $(x + a y)^2 = 4 a^2$ has solutions $x + a y = \pm 2 a$.
- $(a x - y)^2 = a^2$ has solutions $a x - y = \pm a$.
- So the region is bounded by four lines, two parallel pairs.
💡 $X^2 = c$ means $X = \pm\sqrt{c}$ — two parallel lines per equation.
8.EE.B.6 Step 2 - Slopes are perpendicular.
- Solving $x + a y = c$ for $y$ gives slope $-\tfrac{1}{a}$.
- Solving $a x - y = c$ for $y$ gives slope $a$.
- Their product is $a \cdot (-\tfrac{1}{a}) = -1$, so the two pairs are perpendicular.
- Two pairs of perpendicular parallel lines bound a rectangle.
💡 Negative reciprocal slopes $\Rightarrow$ perpendicular lines $\Rightarrow$ region is a rectangle.
8.G.B.7 Step 3 - Distance between parallel lines $x + a y = 2a$ and $x + a y = -2 a$.
- The formula $\tfrac{|c_1 - c_2|}{\sqrt{A^2 + B^2}}$ for $A x + B y = c$ gives length $= \tfrac{|2 a - (-2 a)|}{\sqrt{1^2 + a^2}} = \tfrac{4a}{\sqrt{a^2 + 1}}$.
💡 Distance between two parallel lines $A x + B y = c_1, c_2$ is $|c_1 - c_2| / \sqrt{A^2 + B^2}$ — a Pythagorean-flavored normal projection.
8.G.B.7 Step 4 - Distance between parallel lines $a x - y = a$ and $a x - y = -a$.
- Same formula: width $= \tfrac{|a - (-a)|}{\sqrt{a^2 + (-1)^2}} = \tfrac{2a}{\sqrt{a^2 + 1}}$.
💡 Same formula, this time with $|c_1 - c_2| = 2a$ and the same denominator $\sqrt{a^2 + 1}$.
8.EE.A.2 Step 5 - Area of the rectangle.
- $\text{Area} = \text{length} \times \text{width} = \tfrac{4 a}{\sqrt{a^2 + 1}} \cdot \tfrac{2 a}{\sqrt{a^2 + 1}} = \tfrac{8 a^2}{a^2 + 1}$.
💡 Multiply the two distances; the $\sqrt{a^2 + 1}$'s combine to remove the radical.
8.F.A.3 Step 6 - Spot-check with the easier case $a = 1$.
- The lines become $x + y = \pm 2$ and $x - y = \pm 1$.
- These form a rectangle with length $= \tfrac{4}{\sqrt{2}} = 2\sqrt{2}$ and width $= \tfrac{2}{\sqrt{2}} = \sqrt{2}$, area $= 2\sqrt{2} \cdot \sqrt{2} = 4$.
- The formula gives $\tfrac{8 \cdot 1}{1 + 1} = 4$.
- \checkmark.
- Compare with other choices at $a = 1$: (A) $\tfrac{8}{4} = 2$, (B) $\tfrac{4}{2} = 2$, (C) $\tfrac{8}{2} = 4$, (D) $4$, (E) $\tfrac{8}{2} = 4$ — three choices tie.
- Try $a = 2$: our formula gives $\tfrac{32}{5} = 6.4$.
- Direct rectangle: length $= \tfrac{8}{\sqrt{5}}, $ width $= \tfrac{4}{\sqrt{5}}$, area $= \tfrac{32}{5}$.
- \checkmark.
- Choice (C) at $a = 2$: $\tfrac{16}{3} \approx 5.33$ — different.
- Choice (E) at $a = 2$: $\tfrac{16}{5} = 3.2$ — different.
- Only (D) matches at both values.
💡 Two test values ($a = 1, 2$) suffice to discriminate among the candidate formulas.
8.EE.A.2 Factor each equation into two parallel lines. $(x + a y)^2 = 4 a^2$ has solution 8.EE.B.6 Slopes are perpendicular. Solving $x + a y = c$ for $y$ gives slope $-\tfrac{1}{ 8.G.B.7 Distance between parallel lines $x + a y = 2a$ and $x + a y = -2 a$. The formula 8.G.B.7 Distance between parallel lines $a x - y = a$ and $a x - y = -a$. Same formula: 8.EE.A.2 Area of the rectangle. $\text{Area} = \text{length} \times \text{width} = \tfrac 8.F.A.3 Spot-check with the easier case $a = 1$. The lines become $x + y = \pm 2$ and $x Review
Reasonableness: Behavior as $a \to 0^+$: length $\to 0$ (lines $x + 0 \cdot y = 0$ collapse to $x = 0$), width $\to 0$, area $\to 0$. Our formula: $\tfrac{8 a^2}{a^2 + 1} \to 0$. \checkmark. As $a \to \infty$: length $\approx \tfrac{4 a}{a} = 4$, width $\approx \tfrac{2 a}{a} = 2$, area $\to 8$. Our formula: $\tfrac{8 a^2}{a^2 + 1} \to 8$. \checkmark. Maximum at $a = ?$: derivative shows monotonic increase from $0$ to $8$, never exceeding $8$. Everything geometric is consistent with choice (D).
Alternative: Tool #3 (Eliminate by plugging $a = 1$): the four lines $x + y = \pm 2, x - y = \pm 1$ form a rectangle of area $4$ (easily computed by Shoelace on vertices $(\tfrac{3}{2}, \tfrac{1}{2}), (\tfrac{1}{2}, \tfrac{3}{2}), (-\tfrac{3}{2}, -\tfrac{1}{2}), (-\tfrac{1}{2}, -\tfrac{3}{2})$ found by intersecting). Then test $a = 2$ to break the tie among answers that gave $4$. This route avoids the distance-between-parallel-lines formula entirely if you forgot it.
CCSS standards used (min grade 8)
8.EE.A.2Use square root and cube root symbols to represent solutions (Factoring $X^2 = c^2$ into $X = \pm c$ and writing the area $\tfrac{8a^2}{a^2 + 1}$.)8.EE.B.6Use similar triangles to explain why the slope is the same between any two points (Reading slopes from $y = -\tfrac{1}{a} x + \tfrac{c}{a}$ and $y = a x - c$ and applying the perpendicular slope criterion.)8.F.A.3Interpret the equation $y = m x + b$ as defining a linear function (Sanity-checking the area formula at $a = 1$ and $a = 2$ by graphing the four lines and computing rectangle dimensions.)8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Computing distance between parallel lines via $|c_1 - c_2|/\sqrt{A^2 + B^2}$, which arises from a normal-direction projection (Pythagorean).)
⭐ This AMC 10 problem only needs Grade 8 lines and the Pythagorean distance-between-parallel-lines formula you already know — factor $(X)^2 = c^2$ into two parallel lines per equation, notice the two pairs of slopes $-\tfrac{1}{a}$ and $a$ are perpendicular (so the region is a rectangle), then multiply the two distances $\tfrac{4a}{\sqrt{a^2 + 1}} \cdot \tfrac{2a}{\sqrt{a^2 + 1}} = \tfrac{8a^2}{a^2 + 1}$.
⭐ This AMC 10 problem only needs Grade 8 lines and the Pythagorean distance-between-parallel-lines formula you already know — factor $(X)^2 = c^2$ into two parallel lines per equation, notice the two pairs of slopes $-\tfrac{1}{a}$ and $a$ are perpendicular (so the region is a rectangle), then multiply the two distances $\tfrac{4a}{\sqrt{a^2 + 1}} \cdot \tfrac{2a}{\sqrt{a^2 + 1}} = \tfrac{8a^2}{a^2 + 1}$.