# $a^x+b^x=c^x$ in geometry

• The Pythagorean theorem.
• Let $A$, $C$, $B$ be three points on a line in this order, and let $D$ be another point, such that $\angle ADC =\angle CDB = 60^\circ$. Let $a=AD$, $b=BD$, $c=CD$. Then, $$a^{-1} + b^{-1} = c^{-1}.$$
• Let $C_1$, $C_2$, $C_3$ be three circles that are tangent to each other and also tangent to a common line, such that $C_3$ lies between $C_1$ and $C_2$. Let $a$, $b$, $c$ be their respective radii. Then, $$a^{-1/2} + b^{-1/2} = c^{-1/2}.$$

See the figure below.

Are there any other results of this type in geometry?

• laughs internally Mar 31, 2016 at 18:02
• where are the points $$A,B,C,D$$? Mar 31, 2016 at 18:03
• A rather trivial one: if $A,B,C$ lie on a line in this order and $AB=a,BC=b,CA=c$, then $a^1+b^1=c^1$. Mar 31, 2016 at 18:09
• :) I was waiting for someone to say that... Mar 31, 2016 at 18:09
• That's not exactly related. I'm not requiring a,b,c to be integers, and I want a^x+b^x=c^x to be a theorem related to some (geometrical) construction. Mar 31, 2016 at 18:38

Here's another one. Let $ABCD$ be a trapezoid of area $c$ and let $O$ be the intersection point of its diagonals. If $a$ and $b$ are areas of triangles $AOD$ and $BOC$ (or $AOB$ and $COD$) then $$a^{1/2}+b^{1/2}=c^{1/2}.$$

• The area equation is equivalent to ABCD being a trapezoid.
– zyx
Apr 4, 2016 at 0:35

Exponents $1,2, -1, -2,1/2$, and $-1/2$ are probably the most common, with the complexity of the implied polynomial equation $P(a,b,c)=0$ increasing in that order. As a random example, the height in the crossed ladders problem satisfies $h^{-1} = a^{-1} + b^{-1}$. For the altitude of a right triangle, $h^{-2} = a^{-2} + b^{-2}$. There must be many more.

Exponent $2/3$: the envelope of a line segment of length $L$ with endpoints on the $x$ and $y$ axes is $x^{2/3} + y^{2/3} = L^{2/3}$. The curve is called astroid.

• Nice. Whenever you recall any other examples, put them in. Apr 3, 2016 at 14:13
• The last example generalizes somewhat: the envelope of lines whose x and y intercepts satisfy $a^u + b^u = L^u$ is the curve $x^p+y^p=L^p$ for $p$ Hoelder dual to $-u$. Another geometrically meaningful case is when $a+b=L$ so that the dual is $\sqrt{x}+\sqrt{y}=\sqrt{L}$.
– zyx
Apr 4, 2016 at 0:14

Here's another one. I don't think it's much popular, but the proof is not too terrible.

$$\large a^{2/3}+b^{2/3}=c^{2/3}$$

• So what's the proof? Dec 13, 2021 at 6:21