Find the number of positive integers $b$

Question

Let $a, b$ be positive real numbers such that $10 < a < b$. Then find the number of positive integers $b$ such that (i) $10, a, b$ are in geometric progression, and (ii) $10, a, b$ form the sides of a triangle.

My approach: $10, a, b$ need to satisfy the following conditions:

(i) $10+a>b$

(ii) $a+b>10$

(iii) $10+b>a$

(iv) $a=10r, b=10r^2$ where $r\in (1,\infty)$

using (iv) in the above equations I got:

$r^2-r-1<0$

$r^2+r-1>0$

$r^2-r+1>0$

Then I could not proceed. Help required.

Answer 1

Note that (ii) and (iii) follow from $10<a<b$.

(i) and (iv) together give the inequality you listed as $r^2-r-1<0$.

We can easily solve this using the quadratic formula to give $\frac{1-\sqrt{5}}{2}<r<\frac{1+\sqrt{5}}{2}$.

Using the upper value, we have $b<10 \times (\frac{1+\sqrt{5}}{2})^2\approx 26.18$.

Hence $b$ can be any integer from $11$ to $26$ (inclusive), a total of $16$ possibilities.

Answer 2

The first of the inequalities at the end is the one you want. That gives a limit on $r$. Now try all the $a$ that satisfy $1 \lt r \lt $ this limit. There are not many possibilities-check if $b$ is a natural and you are done.

Answer 3

As has been said, by solving the first inequality and r>1 you can find a range of values that satisfy the criteria. Obviously, you can have many solutions. For simplicity, you can assume the triangle to be right angled and solve.

Looks, like I got the whole thing wrong. Now, for $b to be an integer, $r has to be an integer, which means $a also has to be an integer. The integer greater than one, two, fails with the conditions, which means there is no integral solution to this problem.