# Minimal perimeter

The problem is: Find the angle to OX axis, at which a line should be drawn through the point A (a,b) (a>0, b>0), so, that triangle, formed by this line and positive coordinate semi-axes had the minimal perimeter. I.e. the triangle vertices are (0,0), and two intersection points of line passing through A with OX and OY axes: (x,0), x>0 and (0,y),y>0.

I found the function appearance: the function to minimize is $$f(\varphi)=\left({a\over\cos \varphi}+{b\over\sin \varphi}\right)(1+\cos \varphi+\sin \varphi)$$ and found its derivative,but I failed to solve equation derivative=0: there is 4th degree equation with respect to $\sin\varphi$, $\cos\varphi$ which I failed to solve. The answer is known, it was in the same book, but I can't come to that answer. Could somebody, please, help me?

• It is not clear from this description which triangle you are considering. Where do the three corner points lie? Nov 19, 2016 at 1:18
• one corner point is at (0,0), another on OX axis (i.e. (x,0), x>0), the third is on OY axis (i.e. (0,y),y>0). Nov 19, 2016 at 2:22
• Was the answer something like $\pi/4 \pm \widehat{OO'P}$? See Triangle of Minimum Perimeter for an entertaining discussion of the (not so obvious) geometric approach.
– dxiv
Nov 19, 2016 at 2:41
• As I remember, the answer is $\arcsin{a\over\sqrt{a^2+b^2}}-\arcsin{a-b\over\sqrt{a^2+b^2}}$. Nov 19, 2016 at 2:50

I got the same function $f(\varphi)$ with $0<\varphi<\frac{\pi}{2}$, so I tried to derivate and to solve the equation:

$$f'(\varphi)=\left(a\frac{\sin\varphi}{\cos^2\varphi}-b\frac{\cos\varphi}{\sin^2\varphi}\right)(1+\cos\varphi+\sin\varphi)+\left(\frac{a}{\cos\varphi}+\frac{b}{\sin\varphi}\right)(-\sin\varphi+\cos\varphi)=\\=\left(a\frac{\sin\varphi}{\cos^2\varphi}-b\frac{\cos\varphi}{\sin^2\varphi}\right)(1+\cos\varphi+\sin\varphi)+\left(\frac{a}{\cos\varphi}+\frac{b}{\sin\varphi}\right)(-\sin\varphi+\cos\varphi)=\\=a\frac{\sin\varphi}{\cos^2\varphi}-b\frac{\cos\varphi}{\sin^2\varphi}+a\frac{\sin\varphi}{\cos\varphi}-b\frac{\cos^2\varphi}{\sin^2\varphi}+a\frac{\sin^2\varphi}{\cos^2\varphi}-b\frac{\cos\varphi}{\sin\varphi}-a\frac{\sin\varphi}{\cos\varphi}+a-b+b\frac{\cos\varphi}{\sin\varphi}\\=a\frac{\sin\varphi}{\cos^2\varphi}-b\frac{\cos\varphi}{\sin^2\varphi}-b\frac{\cos^2\varphi}{\sin^2\varphi}+a\frac{\sin^2\varphi}{\cos^2\varphi}+a-b=\\=\frac{a\sin^3\varphi-b\cos^3\varphi-b\cos^4\varphi+a\sin^4\varphi+a\sin^2\varphi\cos^2\varphi-b\sin^2\varphi\cos^2\varphi}{\sin^2\varphi\cos^2\varphi}=\\=\frac{a\sin^2\varphi(\sin^2\varphi+\sin\varphi+\cos^2\varphi)-b\cos^2\varphi(\cos^2\varphi+\cos\varphi+\sin^2\varphi)}{\sin^2\varphi\cos^2\varphi}=\\=\frac{a\sin^2\varphi(1+\sin\varphi)-b\cos^2\varphi(1+\cos\varphi)}{\sin^2\varphi\cos^2\varphi}$$

Putting it equal zero:

$$\frac{a\sin^2\varphi(1+\sin\varphi)-b\cos^2\varphi(1+\cos\varphi)}{\sin^2\varphi\cos^2\varphi}=0\\a\sin^2\varphi(1+\sin\varphi)-b\cos^2\varphi(1+\cos\varphi)=0$$

Now we can use $\sin\varphi=\frac{2t}{1+t^2}$ and $\cos\varphi=\frac{1-t^2}{1+t^2}$ with $t=\tan\frac{\varphi}{2}, \varphi\ne\pi+2k\pi$:

$$a\frac{4t^2}{(1+t^2)^2}\left(1+\frac{2t}{1+t^2}\right)-b\frac{(1-t^2)^2}{(1+t^2)^2}\left(1+\frac{1-t^2}{1+t^2}\right)=0\\4at^2\left(\frac{1+t^2+2t}{1+t^2}\right)-b(1-t^2)^2\left(\frac{1+t^2+1-t^2}{1+t^2}\right)=0\\4at^2\frac{(t+1)^2}{1+t^2}-b(1-t^2)^2\left(\frac{2}{1+t^2}\right)=0\\4at^2(t+1)^2-2b(1-t^2)^2=0\\2at^2(t+1)^2-b(1-t)^2(1+t)^2=0\\2at^2-b(1-t)^2=0\\2at^2-b(1+t^2-2t)=0\\(2a-b)t^2+2bt-b=0$$

Solving this equation I got that the minimum is:

$$t=\frac{-b+\sqrt{2ab}}{2a-b}\\\tan\frac{\varphi}{2}=\frac{-b+\sqrt{2ab}}{2a-b}$$

so $\varphi=2\arctan\frac{-b+\sqrt{2ab}}{2a-b}$.

This is my solution but if someone checked it I would be grateful; I hope it is helpful anyway.

• You are right, although in the book the answer is in another form, I checked, they are equal. Here is how to get the answer from the book (I finally got it with your help!): Nov 19, 2016 at 20:48
• From your last equation: $$a\sin^2\varphi(1+\sin\varphi)-b\cos^2\varphi(1+\cos\varphi)=a(1-\cos^2\varphi)(1+\sin\varphi)-b(1-\sin^2\varphi)(1+\cos\varphi)=a(1-\cos\varphi)-b(1-\sin\varphi)=0$$ $$a\cos\varphi-b\sin\varphi=a-b$$ $${a\over\sqrt{a^2+b^2}}\cos\varphi-{b\over\sqrt{a^2+b^2}}\sin\varphi={a-b\over\sqrt{a^2+b^2}}$$ $$\varphi=\arcsin{a\over\sqrt{a^2+b^2}}-\arcsin{a-b\over\sqrt{a^2+b^2}}$$ Nov 19, 2016 at 21:10
• Yes it is right!! Good job;-) Nov 19, 2016 at 21:46