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Chapter 13: Problem 28

The lengths of the sides in a right triangle form three consecutive terms of a geometric sequence. Find the common ratio of the sequence. (There are two distinct answers.)

Short Answer

Expert verified
The common ratios are \( r = \sqrt{\frac{1 + \sqrt{5}}{2}} \) and \( r = \frac{1}{\sqrt{\frac{1 + \sqrt{5}}{2}}} \).

Step by step solution

01

Identify Triangle Properties

In a right triangle, by the Pythagorean theorem, if the sides are labeled as \( a \), \( b \), and \( c \) (with \( c \) being the hypotenuse), then it holds that \( a^2 + b^2 = c^2 \). In this problem, the sides follow a geometric sequence pattern, meaning the elements are \( a \), \( ar \), and \( ar^2 \) for some common ratio \( r \).
02

Set up the Equation

Assuming \( ar^2 \) is the hypotenuse (\( c \)), thus: \( a = x, \) \( b = ar = xr = yr, \) and \( c = ar^2 = xr^2 = z \). The equation by the Pythagorean theorem becomes: \( x^2 + (xr)^2 = (xr^2)^2 \).
03

Simplify the Equation

Plugging into the Pythagorean theorem: \( x^2 + x^2r^2 = x^2r^4 \). Simplifying this, we get \( 1 + r^2 = r^4 \).
04

Solve the Equation

The equation \( 1 + r^2 = r^4 \) can be rearranged to the polynomial \( r^4 - r^2 - 1 = 0 \). Let \( u = r^2 \), then the equation becomes \( u^2 - u - 1 = 0 \).
05

Factor or Use the Quadratic Formula

Solve \( u^2 - u - 1 = 0 \) using the quadratic formula: \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 1 \), \( b = -1 \), and \( c = -1 \), so \( u = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2} \).
06

Determine Values of \( r \)

Since \( u = r^2 \), take the square roots of the solutions for \( u \): \( r = \pm \sqrt{\frac{1 + \sqrt{5}}{2}} \) and \( r = \pm \sqrt{\frac{1 - \sqrt{5}}{2}} \). However, only positive values are valid for lengths, leading to \( r = \sqrt{\frac{1 + \sqrt{5}}{2}} \) and \( r = \sqrt{\frac{1 - \sqrt{5}}{2}} \), but \( \frac{1 - \sqrt{5}}{2} \) is negative, and so is impractical for \( r \).
07

Conclusion

Thus, the two practical positive values for \( r \) when dealing with side lengths are \( \sqrt{\frac{1 + \sqrt{5}}{2}} \) and \( \frac{1}{\sqrt{\frac{1 + \sqrt{5}}{2}}} \). These lead to different sequences with increasing and decreasing sides, respectively.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Geometric Sequence
A geometric sequence is a sequence of numbers where each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio. In the context of a right triangle, if the sides are part of a geometric sequence, it means the lengths follow this consistent multiplying pattern:
  • The first side is denoted as \( a \).
  • The second side, \( b \), is the first side multiplied by the common ratio \( r \) (hence, \( ar \)).
  • The third side, \( c \), is the second side multiplied again by \( r \) (resulting in \( ar^2 \)).
This setup helps in simplifying complex relationships in right triangles by connecting the sides in a systematic sequence that underscores proportionality across the triangle's dimensions.
Pythagorean Theorem
The Pythagorean theorem is a fundamental principle in geometry, specifically pertinent to right triangles. It states that in a right triangle, the sum of the squares of the two shorter sides is equal to the square of the longest side, known as the hypotenuse. Mathematically, this is expressed as:
  • \( a^2 + b^2 = c^2 \)
In the given problem, the sides conform to a geometric sequence. Assigning the terms \( a, ar, ar^2 \) to the triangle:
  • \( a \) represents the first side.
  • \( ar \) represents the second side.
  • \( ar^2 \) represents the hypotenuse.
Using the theorem, you substitute these expressions into \( a^2 + b^2 = c^2 \) which provides the essential equation to solve for \( r \), the common ratio.
Quadratic Formula
The quadratic formula is a powerful tool used to solve quadratic equations, which are polynomials of degree two. Equations of this type have the general form \( ax^2 + bx + c = 0 \). The formula to find the solutions is:
  • \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
In the context of the exercise, we derive an equation \( u^2 - u - 1 = 0 \) for the square of the common ratio \( r \) (where \( u = r^2 \)). With coefficients \( a = 1 \), \( b = -1 \), and \( c = -1 \), we use the quadratic formula to find \( u \):
  • \( u = \frac{1 \pm \sqrt{5}}{2} \)
Once these solutions for \( u \) are found, it's necessary to determine \( r \) by taking the square root of the positive result, since side lengths cannot be negative. This forms the basis of finding viable geometric sequence terms for the triangle's sides.

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Most popular questions from this chapter

Two real numbers \(A\) and \(B\) are defined by \(A=\sqrt[99]{99 !}\) and \(B=\sqrt[100]{100 !} .\) Which number is larger, \(A\) or \(B ?\) Hint: Compare \(A^{9900}\) and \(B^{9900}\).

Carry out the indicated expansions. $$(x+\sqrt{2})^{8}$$

Find the indicated roots. Express the results in rectangular form. If \(w+\frac{1}{w}=2 \cos \theta,\) show that \(w=\cos \theta \pm i \sin \theta\) Hint: Use the quadratic formula to solve the given equation for \(w\)

Rewrite the sums using sigma notation. $$2-2^{2}+2^{3}-2^{4}+2^{5}$$

A sequence is defined recursively as follows: \(s_{1}=0.7\) and \(s_{n}=\left(s_{n-1}\right)^{2}\) for \(n \geq 2\) (a) Compute the first six terms of this sequence. (Use a calculator; for the answers, round to five decimal places.) What do you observe about the answers? (b) Use a calculator to compute \(s_{10}\). (Report the answer as shown on your calculator screen.) (c) In view of your work in parts (a) and (b), what number do you think would be a very close approximation to \(s_{100} ?\)
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