2021 AMC 10 A Problem 24

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In This Page:

To help you master 2021 AMC 10 A Problem 24, this page includes:

The original AMC problem to try, plus the correct answer

Clear, step-by-step solutions using different methods

Key ideas, useful formulas, and takeaways you can apply to similar problems

Related problems from past AMC exams, plus new practice problems that build the same skills

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2021 AMC10 A Problem 24

Solution

Solution 1

Geometric Distance Method

Taking the square root of the given equations reveals that the region is bounded by two pairs of parallel lines. Because the two pairs are perpendicular to each other, the bounded quadrilateral is a rectangle. We can find its area by calculating its length and width, which are simply the distances between the parallel lines.

To find the distance $d$ between two parallel lines given by $A x + B y = C_{1}$ and $A x + B y = C_{2}$ , use the formula $d = \frac{∣ C _{1} - C _{2} ∣}{A ^{2} + B ^{2}}$ .

Step 1: Simplify the boundary equations

Recognize that both given equations are perfect squares. Taking the square root of both sides reveals the linear equations of the four boundary lines. Be sure to include both positive and negative roots.

(x + a y)^{2} = 4 a^{2} ⟹ x + a y = \pm 2 a

(a x - y)^{2} = a^{2} ⟹ a x - y = \pm a

Step 2: Determine the shape of the bounded region

The four equations represent two pairs of parallel lines. To find the angle between these pairs, examine their slopes by rewriting them in slope-intercept form. Since the product of their slopes is $- 1$ , the adjacent sides are perpendicular. A parallelogram with perpendicular sides is a rectangle.

x + a y = C ⟹ y = - \frac{1}{a} x + \frac{C}{a} (Slope m_{1} = - \frac{1}{a})

a x - y = D ⟹ y = a x - D (Slope m_{2} = a)

m_{1} \cdot m_{2} = (- \frac{1}{a}) (a) = - 1

Step 3: Calculate the side lengths

The side lengths of the rectangle are simply the perpendicular distances between the parallel lines. Apply the formula for the distance between two parallel lines.

d_{1} = \frac{∣2 a - ( - 2 a ) ∣}{1 ^{2} + a ^{2}} = \frac{4 a}{a ^{2} + 1}

d_{2} = \frac{∣ a - ( - a ) ∣}{a ^{2} + ( - 1 ) ^{2}} = \frac{2 a}{a ^{2} + 1}

Step 4: Calculate the area of the rectangle

Multiply the length and width to find the area of the rectangle.

Area = d_{1} \times d_{2}

Area = \frac{4 a}{a ^{2} + 1} \times \frac{2 a}{a ^{2} + 1} = \frac{8 a ^{2}}{a ^{2} + 1}

Correct answer: D. $\frac{8 a ^{2}}{a ^{2} + 1}$
The area is

\frac{8 a ^{2}}{a ^{2} + 1}

, which matches choice D. As a magnitude estimation check, consider what happens as

a \to \infty

. The first equation

(x / a + y)^{2} = 4

forces

y \approx \pm 2

, and the second equation

(a x - y)^{2} = a^{2} ⟹ (x - y / a)^{2} = 1

forces

x \approx \pm 1

. This outlines a

2 \times 4

rectangle with an area of

8

. Our formula gives

lim_{a \to \infty} \frac{8 a ^{2}}{a ^{2} + 1} = 8

, which perfectly confirms our result!

Solution 2

Linear Transformation Method

The expressions $(x + a y)$ and $(a x - y)$ inside the squares suggest a natural change of variables. By defining new variables $u$ and $v$ , we map the slanted, complex quadrilateral in the $x y$ -plane to a simple rectangle in the $uv$ -plane. The area of the original shape is the area of the new shape divided by the transformation’s determinant.

When a region in the $x y$ -plane is transformed into the $uv$ -plane by a linear substitution defined by a matrix $M$ , the area in the new system is scaled by the absolute value of the determinant of $M$ . Thus, $Area_{uv} = ∣ det M ∣ \times Area_{x y}$ .

Step 1: Define a linear coordinate transformation

Let $u$ and $v$ be the expressions inside the squares. This changes our coordinate system from $(x, y)$ to $(u, v)$ .

u = x + a y

v = a x - y

Step 2: Find the area in the new coordinate system

Substitute $u$ and $v$ into the original inequalities to find the boundaries of the region in the $uv$ -plane. This forms a rectangle centered at the origin. Compute its area.

u^{2} \leq 4 a^{2} ⟹ - 2 a \leq u \leq 2 a (Width = 4 a)

v^{2} \leq a^{2} ⟹ - a \leq v \leq a (Height = 2 a)

Area_{uv} = 4 a \times 2 a = 8 a^{2}

Step 3: Calculate the transformation scaling factor

Extract the coefficient matrix $M$ that transforms $(x, y)$ to $(u, v)$ . The area scaling factor is the absolute value of the determinant of this matrix.

(u v) = (1 a a - 1) (x y)

det M = (1) (- 1) - (a) (a) = - 1 - a^{2}

∣ det M ∣ = ∣ - 1 - a^{2} ∣ = a^{2} + 1

Step 4: Compute the original area

The area in the $uv$ -plane is equal to the original area in the $x y$ -plane multiplied by the scaling factor. Rearrange to solve for the target area.

Area_{uv} = ∣ det M ∣ \times Area_{x y}

8 a^{2} = (a^{2} + 1) \times Area_{x y}

Area_{x y} = \frac{8 a ^{2}}{a ^{2} + 1}

Correct answer: D. $\frac{8 a ^{2}}{a ^{2} + 1}$
The area is

\frac{8 a ^{2}}{a ^{2} + 1}

, which matches choice D. The determinant method completely sidesteps geometric distance calculations and provides the same result, confirming our previous work.

Solution 3

Multiple-Choice Shortcut

Since the problem asks for an expression valid for all positive real numbers $a$ , we can plug in specific values for $a$ to calculate a numerical area. We then evaluate the answer choices using the same $a$ and eliminate any options that do not match our result.

When a problem guarantees that an algebraic formula holds true for all valid variables, substituting a specific, convenient number is a fast and rigorous way to eliminate incorrect multiple-choice options.

Step 1: Test a simple value for $a$

Substitute $a = 1$ into the original equations to simplify them.

(x + y)^{2} = 4 ⟹ x + y = \pm 2

(x - y)^{2} = 1 ⟹ x - y = \pm 1

Step 2: Calculate the area for $a = 1$

Find the distances between these parallel lines and multiply them to get the area of the resulting rectangle.

d_{1} = \frac{∣2 - ( - 2 ) ∣}{1 ^{2} + 1 ^{2}} = \frac{4}{2} = 22

d_{2} = \frac{∣1 - ( - 1 ) ∣}{1 ^{2} + ( - 1 ) ^{2}} = \frac{2}{2} = 2

Area = 22 \times 2 = 4

Step 3: Evaluate the answer choices for $a = 1$

Substitute $a = 1$ into the five options to see which ones produce an area of $4$ . This eliminates A and B.

A. \frac{8 ( 1 ) ^{2}}{( 1 + 1 ) ^{2}} = 2 B. \frac{4 ( 1 )}{1 + 1} = 2

C. \frac{8 ( 1 )}{1 + 1} = 4 D. \frac{8 ( 1 ) ^{2}}{1 ^{2} + 1} = 4 E. \frac{8 ( 1 )}{1 ^{2} + 1} = 4

Step 4: Test a second value to break the tie

Because choices C, D, and E all tied at $4$ , we test $a = 2$ to isolate the correct option.

(x + 2 y)^{2} = 16 ⟹ x + 2 y = \pm 4

(2 x - y)^{2} = 4 ⟹ 2 x - y = \pm 2

d_{1} = \frac{∣4 - ( - 4 ) ∣}{1 ^{2} + 2 ^{2}} = \frac{8}{5}

d_{2} = \frac{∣2 - ( - 2 ) ∣}{2 ^{2} + ( - 1 ) ^{2}} = \frac{4}{5}

Area = \frac{8}{5} \times \frac{4}{5} = \frac{32}{5} = 6.4

Step 5: Evaluate the remaining choices for $a = 2$

Substitute $a = 2$ into the remaining three choices. Only one will evaluate to $6.4$ .

C. \frac{8 ( 2 )}{2 + 1} = \frac{16}{3} \approx 5.33

D. \frac{8 ( 2 ) ^{2}}{2 ^{2} + 1} = \frac{32}{5} = 6.4

E. \frac{8 ( 2 )}{2 ^{2} + 1} = \frac{16}{5} = 3.2

Correct answer: D. $\frac{8 a ^{2}}{a ^{2} + 1}$
Evaluating the formulas for

a = 2

proves that choice D is the only expression that produces the correct numerical area of

6.4

.

Adapted Problems to Practice

1 / 10

What Formulas Do You Need for This Problem?

Distance Between Parallel Lines:

d = \frac{∣ C _{1} - C _{2} ∣}{A ^{2} + B ^{2}}

$A$ , $B$ , $C_{1}$ , and $C_{2}$ are constants from the equations of two parallel lines written in the form $A x + B y = C_{1}$ and $A x + B y = C_{2}$ . We used this formula to find the length and width of the bounding rectangle directly from the equations of its sides.

Perpendicular Lines Rule:

Two lines $A_{1} x + B_{1} y = C$ and $A_{2} x + B_{2} y = D$ are perpendicular if their slopes multiply to $- 1$ . A quick way to recognize perpendicular lines is if their coefficients are swapped and one sign is negated, such as $A x + B y = C$ and $B x - A y = D$ .

Applies to lines in the Cartesian coordinate system. The given equations $x + a y = C$ and $a x - y = D$ fit this exact pattern, immediately verifying that the region is a rectangle rather than a slanted parallelogram.

Key Takeaway: Remember This Strategy for Next Time

When a problem defines boundaries using perfect squares, such as $(e x p r)^{2} = C^{2}$ .

Try: Take the square root of both sides to extract linear equations. Always remember the $\pm$ sign, which naturally yields pairs of parallel lines.

Why: Taking the square root simplifies the geometry from quadratic curves to straightforward straight lines.

When boundary lines take the form $A x + B y = C$ and $B x - A y = D$ .

Try: Immediately recognize that the lines are perpendicular.

Why: The slopes are $- A / B$ and $B / A$ . Multiplying these gives $- 1$ , which is the geometric definition of perpendicular slopes. This guarantees the bounded quadrilateral is a rectangle.

When a multiple-choice geometry or algebra problem features variables in both the problem statement and the answer choices.

Try: Substitute specific, easy-to-compute values (like $1$ and $2$ ) for the variable. Calculate the numerical answer and eliminate any formulas that don't match.

Why: If a generalized formula is correct, it must mathematically hold true for every valid numerical case.

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2021 AMC 10 A Problem 24

In This Page:

2021 AMC10 A Problem 24

Solution

Solution 1

Geometric Distance Method

Step 1: Simplify the boundary equations

Step 2: Determine the shape of the bounded region

Step 3: Calculate the side lengths

Step 4: Calculate the area of the rectangle

Solution 2

Linear Transformation Method

Step 1: Define a linear coordinate transformation

Step 2: Find the area in the new coordinate system

Step 3: Calculate the transformation scaling factor

Step 4: Compute the original area

Solution 3

Multiple-Choice Shortcut

Step 1: Test a simple value for a

Step 2: Calculate the area for a=1

Step 3: Evaluate the answer choices for a=1

Step 4: Test a second value to break the tie

Step 5: Evaluate the remaining choices for a=2

Similar Problems in Real Exams

2021 AMC 10 A Problem 19

2025 AMC 10 A Problem 3

2025 AMC 10 A Problem 6

2025 AMC 10 A Problem 10

2025 AMC 10 A Problem 15

Adapted Problems to Practice

What Formulas Do You Need for This Problem?

Key Takeaway: Remember This Strategy for Next Time

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Step 1: Test a simple value for $a$

Step 2: Calculate the area for $a = 1$

Step 3: Evaluate the answer choices for $a = 1$

Step 5: Evaluate the remaining choices for $a = 2$

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