Question

use the golden rectangles shown to the right to obtain an exact value for \\( \phi \\), the ratio of the long side to the short side in a golden rectangle of any size. complete parts (a) through (c) below.

a. the golden ratio in rectangle a, or the ratio of the long side to the short side, can be modeled by \\( \frac{\phi}{1} \\). which of the following fractional expressions models the golden ratio in rectangle b?
\\( \bigcirc \\) a. \\( \frac{\phi}{\phi - 1} \\) \\( \bigcirc \\) b. \\( \frac{\phi - 1}{1} \\) \\( \bigcirc \\) c. \\( \frac{\phi - 1}{\phi} \\) \\( \bigcirc \\) d. \\( \frac{1}{\phi - 1} )

b. set the expression for the golden ratio in rectangle a equal to the expression for the golden ratio in rectangle b. solve the resulting proportion using the quadratic formula. express \\( \phi \\) as an exact value in simplified radical form.
\\( \phi = \square \\) (type an exact answer)

Explanation:

Part (a)

In a golden rectangle, the ratio of long to short side is consistent. For rectangle A, the ratio is $\frac{\Phi}{1}$. For rectangle B, the long side is $1$ and the short side is $\Phi - 1$, so the ratio is $\frac{1}{\Phi - 1}$? Wait, no—wait, actually, rectangle B has length $\Phi - 1$ and width $1$? Wait, no, looking at the diagram: Golden Rectangle A has length $\Phi$ and width $1$. The square has side $1$, so the remaining rectangle (B) has length $1$ and width $\Phi - 1$? Wait, no, the ratio of long to short in a golden rectangle is the same for A and B. So for rectangle A: long is $\Phi$, short is $1$, ratio $\Phi/1$. For rectangle B: long is $1$, short is $\Phi - 1$? No, wait, no—wait, the golden ratio property is that if you remove a square from a golden rectangle, the remaining rectangle is also a golden rectangle. So the ratio of the original (A) is $\Phi$ (long/short = $\Phi/1$), and the remaining rectangle (B) has long side $1$ and short side $\Phi - 1$, so its ratio should also be $\Phi$. Wait, no, the ratio of long to short for B should be equal to the ratio for A. So for B, long side is $1$, short side is $\Phi - 1$? No, that can't be. Wait, maybe I got the sides reversed. Let's re-examine: The total length of A is $\Phi$, with a square of side $1$ and a smaller golden rectangle B. So the length of B is $\Phi - 1$, and the width of B is $1$ (since the square has side $1$). Wait, no, the height of both A and B is $1$. So for A: length $\Phi$, width $1$, ratio $\Phi/1$. For B: length $1$, width $\Phi - 1$? No, that would make the ratio $1/(\Phi - 1)$, but that should equal $\Phi$ (since it's a golden rectangle). Wait, no, the ratio of long to short in B should be the same as in A. So if A has ratio $\Phi$ (long/short = $\Phi/1$), then B, which is also a golden rectangle, should have long/short = $\Phi$. The long side of B is $1$ (since the square is side $1$, and the height is $1$), and the short side is $\Phi - 1$. Wait, no, maybe the long side of B is $\Phi - 1$ and the short side is $1$? No, that would make the ratio $(\Phi - 1)/1$, but that should equal $\Phi$. Wait, I think I messed up. Let's recall the golden ratio: the ratio of the longer side to the shorter side is equal to the ratio of the sum of the sides to the longer side. So in the diagram, the big rectangle (A) has length $\Phi$ and width $1$. The square is $1 \times 1$, so the smaller rectangle (B) has length $1$ and width $\Phi - 1$. For B to be a golden rectangle, the ratio of its long side to short side should be equal to $\Phi$. So the long side of B is $1$, short side is $\Phi - 1$, so ratio is $1/(\Phi - 1) = \Phi$. Wait, but the options are: A. $\Phi/(\Phi - 1)$, B. $(\Phi - 1)/1$, C. $(\Phi - 1)/\Phi$, D. $1/(\Phi - 1)$. Wait, the question says "the golden ratio in rectangle A, or the ratio of the long side to the short side, can be modeled by $\Phi/1$". Then "Which of the following fractional expressions models the golden ratio in rectangle B?" So for rectangle B, the long side and short side: the height is $1$, the length is $\Phi - 1$? No, wait, the total length of A is $\Phi$, with a square of side $1$, so the remaining length is $\Phi - 1$. So the smaller rectangle B has length $\Phi - 1$ and height $1$? No, that would make the long side $1$ (height) and short side $\Phi - 1$ (length) if $\Phi - 1 < 1$. Wait, $\Phi$ is the golden ratio, approximately $1.618$, so $\Phi - 1 \approx 0.618$, which is less than $1$. So the long side of B is $1$ (height), short side is $\Phi - 1$ (length). Therefore, the ratio of long to short in B is $1/(\Phi - 1)$.…

Step 1: Set up the proportion

From part (a), the ratio for A is $\frac{\Phi}{1}$ and for B is $\frac{1}{\Phi - 1}$. Since they are equal (golden ratio property), we set:
$$\frac{\Phi}{1} = \frac{1}{\Phi - 1}$$

Step 2: Cross - multiply to form a quadratic equation

Cross - multiplying gives:
$$\Phi(\Phi - 1)=1$$
Expand the left - hand side:
$$\Phi^{2}-\Phi = 1$$
Rearrange to standard quadratic form $ax^{2}+bx + c = 0$:
$$\Phi^{2}-\Phi - 1=0$$

Step 3: Apply the quadratic formula

For a quadratic equation $ax^{2}+bx + c = 0$, the quadratic formula is $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Here, $a = 1$, $b=-1$, and $c=-1$. Substitute these values into the formula:
$$\Phi=\frac{-(-1)\pm\sqrt{(-1)^{2}-4(1)(-1)}}{2(1)}$$
Simplify the numerator:
First, $-(-1)=1$. Then, $(-1)^{2}-4(1)(-1)=1 + 4=5$. So:
$$\Phi=\frac{1\pm\sqrt{5}}{2}$$

Step 4: Determine the positive solution

Since $\Phi$ represents a ratio of lengths, it must be positive. We have two solutions: $\frac{1+\sqrt{5}}{2}$ and $\frac{1 - \sqrt{5}}{2}$. The second solution $\frac{1-\sqrt{5}}{2}\approx\frac{1 - 2.236}{2}\approx - 0.618$ is negative, so we discard it. Thus, $\Phi=\frac{1+\sqrt{5}}{2}$.

Answer:

D. $\boldsymbol{\frac{1}{\Phi - 1}}$

Part (b)