Prove that the cube roots of areas are equal.

Question

In $\triangle ABC$, $X$ and $Y$ are points on the sides $AC$ and $BC$ respectively. If $Z$ is on the segment $XY$ such that $\frac{AX}{XC}=\frac{CY}{YB}=\frac{XZ}{ZY}$. Prove that the area of $\triangle ABC$ is given by $\triangle ABC = \left((\triangle AXZ)^{1/3}+(\triangle BYZ)^{1/3}\right)^3$ or alternatively $(\triangle ABC)^{1/3}=(\triangle AXZ)^{1/3}+(\triangle BYZ)^{1/3}$

I tried to use Menelaus theorem a bunch of times to leverage the ratios given but it does not get me anywhere useful so that I can relate them with the areas. I would be grateful if anybody could provide a solution. And preferably one that uses simple methods like similarity, Menelaus theorem. Ceva's theorem, etc., since this question is taken from an exercise on the above-mentioned topics. Of course, other solutions are always welcome, since it is always better to know multiple ways to solve the same problem.

There is a connection 1) with the so called "centro-affine geometry" ; see fig. 3 in this answer of mine to a different question : https://math.stackexchange.com/q/3564572, 2) With Bezier curves. — Jean Marie, Aug 09 '20 at 21:35

score 1 · Accepted Answer · answered Aug 08 '20 at 09:08

1

Let the ratio $AX\over XC$ be $r$.

Then $S_{ACY}={r\over 1+r}S_{ABC}$, $S_{AXY}={r\over 1+r}S_{ACY}$, $S_{AXZ}={r\over 1+r}S_{AXY}$

$$S_{AXZ} = ({r\over 1+r})^3S_{ABC}$$

Similarly, we can find

$$S_{BYZ} = ({1\over 1+r})^3S_{ABC}$$

The equality thus follows.

answered Aug 08 '20 at 09:08

cr001

12,778

From $S$ do you mean area? – Popular Power Aug 08 '20 at 11:50
Yes, $S_{ABC}$ means area of triangle $ABC$. – cr001 Aug 08 '20 at 12:02
1

lol, I cannot imagine why I did not connect those two lines. I am really a fool. – Popular Power Aug 08 '20 at 12:05

Prove that the cube roots of areas are equal.

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