Little and large
A point moves around inside a rectangle. What are the least and the
greatest values of the sum of the squares of the distances from the
vertices?
Age
16 to 18
Challenge level
Problem
Image
The point $X$ moves around inside a rectangle of dimension $p$ units by $q$ units. The distances of $X$ from the vertices of the rectangle are $a$, $b$, $c$ and $d$ units. What are the least and the greatest values of
$a^2 + b^2 + c^2 + d^2$
and where is the point $X$ when these values occur?
Student Solutions
Congratulations to Soh Yong Sheng from Raffles Institution,
Singapore for this excellent solution.
We have $0 < a < b$ which means $1/a > 1/b$ and so $3 + 1/a > 3 + 1/b$. Flipping over the fraction, we will get 1/(3 + 1/a) < 1/(3 + 1/b) and the inequality remains the same way round when 2 is added. Flipping over again for the last time we get \[ \frac{1}{2+\frac{1}{3+\frac{1}{a}}} \] is greater than \[ \frac{1}{2+\frac{1}{3+\frac{1}{b}}} \]
The second part is a further expansion of the first, and in the process of repeating the above we know that it involves just one more flipping over of the fraction, thus \[ \frac{1}{2+\frac{1}{3+\frac{1}{4 + \frac{1}{a}}}} \] is less than the same thing with $b$ in place of a as the inequality would be reversed again.
Lastly the continued fractions are expanded all the way down to $100 + 1/a$ and $100 + 1/b$. Observe the above process, we can tell that if the last or biggest number is odd then the continued fraction with a in it is bigger. If the last or biggest number is even then the continued fraction with $b$ in it is bigger. Each successive continued fraction involves one more 'flipping over' and reverses the inequality one more time. The following continued fraction is smaller than the same thing with $b$ in place of $a$: $${1\over\displaystyle 2 + { 1 \over \displaystyle 3+ { 1\over \displaystyle 4 + \dots + {1\over\displaystyle 99+ {1\over \displaystyle {100 + {1 \over \displaystyle a}} }}}}}$$
We have $0 < a < b$ which means $1/a > 1/b$ and so $3 + 1/a > 3 + 1/b$. Flipping over the fraction, we will get 1/(3 + 1/a) < 1/(3 + 1/b) and the inequality remains the same way round when 2 is added. Flipping over again for the last time we get \[ \frac{1}{2+\frac{1}{3+\frac{1}{a}}} \] is greater than \[ \frac{1}{2+\frac{1}{3+\frac{1}{b}}} \]
The second part is a further expansion of the first, and in the process of repeating the above we know that it involves just one more flipping over of the fraction, thus \[ \frac{1}{2+\frac{1}{3+\frac{1}{4 + \frac{1}{a}}}} \] is less than the same thing with $b$ in place of a as the inequality would be reversed again.
Lastly the continued fractions are expanded all the way down to $100 + 1/a$ and $100 + 1/b$. Observe the above process, we can tell that if the last or biggest number is odd then the continued fraction with a in it is bigger. If the last or biggest number is even then the continued fraction with $b$ in it is bigger. Each successive continued fraction involves one more 'flipping over' and reverses the inequality one more time. The following continued fraction is smaller than the same thing with $b$ in place of $a$: $${1\over\displaystyle 2 + { 1 \over \displaystyle 3+ { 1\over \displaystyle 4 + \dots + {1\over\displaystyle 99+ {1\over \displaystyle {100 + {1 \over \displaystyle a}} }}}}}$$