AMC 10 · 2023 · #11

Grade 8 geometry-2d

pythagorean-theoremarea-rectanglessimilar-triangles convert-to-algebraidentify-subproblems ↑ Prerequisites: pythagorean-theoremarea-rectangles

📏 Long solution 💡 3 insights 📊 Diagram

Problem

A square of area $2$ is inscribed in a square of area $3$ , creating four congruent triangles, as shown below. What is the ratio of the shorter leg to the longer leg in the shaded right triangle?

$\textbf{(A) }\frac15\qquad\textbf{(B) }\frac14\qquad\textbf{(C) }2-\sqrt3\qquad\textbf{(D) }\sqrt3-\sqrt2\qquad\textbf{(E) }\sqrt2-1$

Pick an answer.

(A)

$\frac{1}{5}$

(B)

$\frac{1}{4}$

(C)

$2-\sqrt{3}$

(D)

$\sqrt{3}-\sqrt{2}$

(E)

$\sqrt{2}-1$

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Toolkit + CCSS Solution

Understand

Restated: A square of area $3$ contains a tilted square of area $2$ that touches the outer square once on each side. That tilting cuts off four congruent right triangles from the corners. Find the ratio of the shorter leg to the longer leg of one of those right triangles.

Givens: Outer square has area $3$, so its side length is $\sqrt{3}$; Inner (tilted) square has area $2$, so its side length is $\sqrt{2}$; Each corner of the outer square is cut off by a right triangle whose hypotenuse is a side of the inner square; All four right triangles are congruent; Answer choices: (A) $\tfrac15$, (B) $\tfrac14$, (C) $2-\sqrt3$, (D) $\sqrt3-\sqrt2$, (E) $\sqrt2-1$

Unknowns: The ratio shorter leg / longer leg of one of the four congruent corner right triangles

Understand

Plan

Primary tool: #1 Draw a Diagram

Secondary: #13 Convert to Algebra, #3 Eliminate Possibilities

Start with Tool #1 (Draw a Diagram): a quick labeled sketch shows that along any side of the outer square the two legs of one corner triangle sit end-to-end, so leg + leg equals the outer side $\sqrt{3}$. The same picture identifies the hypotenuse with a side of the inner square, $\sqrt{2}$, so the Pythagorean theorem gives the second equation. Tool #13 (Convert to Algebra) then turns those two pictorial facts into a clean two-equation system in $x$ (short leg) and $y$ (long leg), which solves to a numerical ratio. Finally Tool #3 (Eliminate Possibilities) double-checks the closed-form answer numerically against the five choices.

Execute — Answer: C

#1 Draw a Diagram 8.EE.A.2 Step 1

Read the side lengths off the areas.
A square's side is the square root of its area, so the outer side is $\sqrt{3}$ and the inner side is $\sqrt{2}$.

$$s_{\text{out}} = \sqrt{3}, \quad s_{\text{in}} = \sqrt{2}$$

💡 Area equals side squared, so taking a square root reverses it — Grade 8 makes the $\sqrt{\,}$ symbol official.

#1 Draw a Diagram 6.EE.A.2 Step 2

Draw and label one corner triangle.
Call the shorter leg $x$ and the longer leg $y$.
The two legs of one corner triangle lie along one side of the outer square, end to end, so $x + y$ equals that side.
The hypotenuse is a side of the inner (tilted) square.

$$x + y = \sqrt{3}, \quad \text{hypotenuse} = \sqrt{2}$$

💡 A picture turns geometry into named lengths; once the legs have names, the relationships become Grade 6 expressions.

#13 Convert to Algebra 8.G.B.7 Step 3

Use the Pythagorean theorem on the right triangle: legs $x$ and $y$, hypotenuse $\sqrt{2}$.

$$x^2 + y^2 = (\sqrt{2})^2 = 2$$

💡 Grade 8 Pythagorean theorem ties the squares of the two legs to the square of the hypotenuse — exactly the bridge from "corner triangle" to a usable equation.

#13 Convert to Algebra 7.EE.A.1 Step 4

Combine the two equations.
Square the first equation to bring it onto the same footing as the second, then subtract to isolate $2xy$.

$$(x+y)^2 = x^2 + 2xy + y^2 = 3 \;\Rightarrow\; 2xy = 3 - (x^2+y^2) = 3 - 2 = 1$$

💡 Expanding $(x+y)^2$ is the Grade 7 move that links the sum of two numbers to the sum of their squares via their product.

#13 Convert to Algebra 8.EE.A.2 Step 5

Now $x$ and $y$ are the two numbers with sum $\sqrt{3}$ and product $\tfrac{1}{2}$.
By Vieta's, they are the roots of $t^2 - \sqrt{3}\,t + \tfrac{1}{2} = 0$.
Multiply through by $2$ and apply the quadratic formula.

$$2t^2 - 2\sqrt{3}\,t + 1 = 0 \;\Rightarrow\; t = \dfrac{2\sqrt{3} \pm \sqrt{12 - 8}}{4} = \dfrac{2\sqrt{3} \pm 2}{4} = \dfrac{\sqrt{3} \pm 1}{2}$$

💡 Sum + product → quadratic. The two roots come out as conjugate pairs, perfect for naming the short and long legs separately.

#13 Convert to Algebra 7.EE.A.1 Step 6

Identify which root is shorter.
Since $\sqrt{3} \approx 1.732$, the two leg lengths are $\tfrac{\sqrt{3}-1}{2} \approx 0.366$ and $\tfrac{\sqrt{3}+1}{2} \approx 1.366$.
Take the ratio shorter:longer.

$$\dfrac{x}{y} = \dfrac{\tfrac{\sqrt{3}-1}{2}}{\tfrac{\sqrt{3}+1}{2}} = \dfrac{\sqrt{3}-1}{\sqrt{3}+1}$$

💡 Dividing two conjugate expressions is a Grade 7 expression-simplification move.

#13 Convert to Algebra 8.EE.A.2 Step 7

Rationalize the denominator by multiplying top and bottom by $\sqrt{3}-1$.

$$\dfrac{\sqrt{3}-1}{\sqrt{3}+1} \cdot \dfrac{\sqrt{3}-1}{\sqrt{3}-1} = \dfrac{(\sqrt{3}-1)^2}{3-1} = \dfrac{3 - 2\sqrt{3} + 1}{2} = \dfrac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}$$

💡 Multiplying by a conjugate cancels the radical in the denominator — the standard Grade 8 square-root manipulation.

#3 Eliminate Possibilities 8.NS.A.2 Step 8

Match against the choices using Tool #3.
The ratio $2 - \sqrt{3} \approx 2 - 1.732 = 0.268$ matches choice (C).
Quick check on the others: (A) $0.20$, (B) $0.25$, (D) $\sqrt{3}-\sqrt{2} \approx 0.318$, (E) $\sqrt{2}-1 \approx 0.414$ — only (C) lands on $0.268$.

$$2 - \sqrt{3} \approx 0.268 \;\Rightarrow\; \textbf{(C)}$$

💡 Rational approximations of irrational numbers (Grade 8) let you compare the closed-form ratio to the five choices instantly.

[1] #1 8.EE.A.2 Read the side lengths off the areas. A square's side is the square root of its a

[2] #1 6.EE.A.2 Draw and label one corner triangle. Call the shorter leg $x$ and the longer leg

[3] #13 8.G.B.7 Use the Pythagorean theorem on the right triangle: legs $x$ and $y$, hypotenuse

[4] #13 7.EE.A.1 Combine the two equations. Square the first equation to bring it onto the same f

[5] #13 8.EE.A.2 Now $x$ and $y$ are the two numbers with sum $\sqrt{3}$ and product $\tfrac{1}{2

[6] #13 7.EE.A.1 Identify which root is shorter. Since $\sqrt{3} \approx 1.732$, the two leg leng

[7] #13 8.EE.A.2 Rationalize the denominator by multiplying top and bottom by $\sqrt{3}-1$.

[8] #3 8.NS.A.2 Match against the choices using Tool #3. The ratio $2 - \sqrt{3} \approx 2 - 1.7

Review

Reasonableness: Numeric sanity: $x + y = \tfrac{\sqrt{3}-1}{2} + \tfrac{\sqrt{3}+1}{2} = \sqrt{3}$ matches the outer side, and $x^2 + y^2 = \tfrac{(\sqrt{3}-1)^2 + (\sqrt{3}+1)^2}{4} = \tfrac{(4-2\sqrt{3}) + (4+2\sqrt{3})}{4} = \tfrac{8}{4} = 2$ matches the squared inner side. Direction check: shorter / longer must be less than $1$, and $2 - \sqrt{3} \approx 0.268 \in (0,1)$ — good.

Alternative: Tool #6 (Guess and Check) on the five choices: each choice is a candidate ratio $r = x/y$ with $x + y = \sqrt{3}$. From $x = ry$ and $x + y = \sqrt{3}$ you get $y = \tfrac{\sqrt{3}}{1+r}$, then $x^2 + y^2 = y^2(1 + r^2)$ must equal $2$. Testing (C) $r = 2 - \sqrt{3}$: $1 + r = 3 - \sqrt{3}$, $y = \tfrac{\sqrt{3}}{3-\sqrt{3}} = \tfrac{\sqrt{3}(3+\sqrt{3})}{6} = \tfrac{3\sqrt{3}+3}{6} = \tfrac{\sqrt{3}+1}{2}$, and a quick squared sum confirms $2$. Same answer, no quadratic formula needed.

CCSS standards used (min grade 8)

6.EE.A.2 Write, read, and evaluate expressions in which letters stand for numbers (Naming the two legs $x$ and $y$ so that the picture's two relationships ($x+y = \sqrt{3}$ and hypotenuse $= \sqrt{2}$) become written equations.)
7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions (Expanding $(x+y)^2$ to link sum-of-roots and sum-of-squares, and simplifying the conjugate ratio $(\sqrt{3}-1)/(\sqrt{3}+1)$.)
8.EE.A.2 Use square root and cube root symbols to represent solutions (Reading $\sqrt{3}$ and $\sqrt{2}$ as side lengths from given areas, and rationalizing $\dfrac{\sqrt{3}-1}{\sqrt{3}+1}$ to a clean form $2 - \sqrt{3}$.)
8.G.B.7 Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Turning the corner right triangle with hypotenuse $\sqrt{2}$ into the equation $x^2 + y^2 = 2$.)
8.NS.A.2 Use rational approximations of irrational numbers to compare their size (Numerically estimating $2 - \sqrt{3} \approx 0.268$ to match the closed-form ratio against the five answer choices.)

⭐ Once you draw the tilted square and label one corner triangle, two facts pop out: the two legs share one outer side ($x + y = \sqrt{3}$), and the hypotenuse is the inner side ($x^2 + y^2 = 2$). From there, Grade 8 algebra of square roots does the rest.