# Thread: Maximise the lengths of two sides of a triangle within a circle

1. ## Maximise the lengths of two sides of a triangle within a circle

I'm doing a bunch of analysis problems and have come across this one.
I feel quite dumb for this, I'm sure this is a 16-year old teenager level basic calculus problem but can't for the life of me remember how to actually formulate the problem.

I have a triangle within a circle (which I'm calling C) of radius 1 which can be seen in the following diagram.
The angle TSP is a right angle.

What is the maximum of the sum of the lengths TS and SP? Presumably, the points S and T can move about providing S stays on the diametre and T stays on the "rim" of the circle and the angle remains as a right angle.
Ie: Find max(TS+SP).

My instinct tells me I need to formulate this into a problem featuring something I can differentiate. Then I just find where the derivative equals zero and go from there, like I said; basic calculus principles. But I'm not even sure how to formulate it into an equation.

Thanks in advance for any assistance y'all can provide.

2. ## Re: Maximise the lengths of two sides of a triangle within a circle

Set up a coordinate system so that Q and P are on the x-axis and the center of the circle is the origin. Then the circle is given by equation $x^2+ y^2= R^2$ where "R" is the length of a radius of the circle (which you don't give but clearly the distance you want depend on R). Take point S to have coordinates (x,0). Then T has coordinates $(x, \sqrt{R^2- x^2})$ and P has coordinates (R, 0). The distance from S to T is $\sqrt{R^2- x^2}$ and the distance from S to P is R- x. So your total distance is given by $R- x+ \sqrt{R^2- x^2}$.

3. ## Re: Maximise the lengths of two sides of a triangle within a circle

Originally Posted by HallsofIvy
Set up a coordinate system so that Q and P are on the x-axis and the center of the circle is the origin. Then the circle is given by equation $x^2+ y^2= R^2$ where "R" is the length of a radius of the circle (which you don't give but clearly the distance you want depend on R). Take point S to have coordinates (x,0). Then T has coordinates $(x, \sqrt{R^2- x^2})$ and P has coordinates (R, 0). The distance from S to T is $\sqrt{R^2- x^2}$ and the distance from S to P is R- x. So your total distance is given by $R- x+ \sqrt{R^2- x^2}$.
Ah yes, thankyou, that makes a lot of sense.

Following this I let the sum equal f(x) and differentiated, let the derivative equal to zero and obtained the stationary points $x=\pm \frac{1}{\sqrt{2}}$
Then $x=\frac{-1}{\sqrt{2}}$ yields the maximum sum which is $1+\sqrt{2}$. (And of course the other root yields the minimum which is 1.)