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Triangle with sides x,y,z. Three equal length line segments parallel to the sides intersect at point W and are bounded by the triangle.

The lengths of the triangle sides satisfy $x < y < z$. Lines that look parallel indeed are parallel. Triangles that look similar indeed are similar. The three red lines are concurrent (meet at a single point) and congruent (have the same length). What is their length?

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closed as off-topic by JonTheMon, JMP, Ian MacDonald, feelinferrety, BmyGuest Mar 21 '16 at 16:31

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  • $\begingroup$ Do you want their length in terms of x, y, and z? $\endgroup$ – Joe Z. Mar 17 '16 at 16:30
  • $\begingroup$ Yeah, a general solution in terms of x, y, and z. I saw this puzzle posed specifically for a 3:4:6 triangle, and hammered away at it for a while before going to bed. $\endgroup$ – Anon Mar 17 '16 at 16:42
  • $\begingroup$ Nice solutions... So the answer to the 3:4:6 triangle is 8. $\endgroup$ – Gamow Mar 17 '16 at 18:11
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Let $a$ be the proportion of the dimensions of the triangle including the line from $y$ to $z$ to those of the full triangle. Define $b$ and $c$ similarly for the lines from $x$ to $z$, and from $x$ to $y$. Then, from similarity, $ax=by=cz$.

Also, if we look at side $x$, we can see from similarity that the section from the bottom to the $xz$ line has length $bx$, the section from the top to the $xy$ line has length $cx$, and the overlap of those two sections (the section between the two red lines) has length $(1-a)x$. Thus

$bx + cx - (1-a)x = x$

$\implies a+b+c=2$

Substituting $b=\frac{ax}{y}$ and $c=\frac{ax}{z}$, we have

$a+\frac{ax}{y}+\frac{ax}{z}=2$

$\implies a(\frac{yz+xz+xy}{yz})=2$

$\implies ax=\frac{2xyz}{xy+yz+zx}$

Since the three lines are congruent, this is the length of all of them.

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  • 1
    $\begingroup$ Duuude...spoilers? :) $\endgroup$ – Marius Mar 17 '16 at 17:58
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Answer

$\frac{2*x*y*z}{x*y + x*z + y*z}$

Proof: Let's convene to the notations on the image:

And let's make the length of the red lines $q$.

$AxH = AyA$
$BxH = BzB$
This means $AB = AxBx + AyBz$ which translates to $AyBz = x-q$
In a similar logic $AxCz = y-q$
Triangle AyBzH is similar to ABC This results in $\frac{AyBz}{x} = \frac{BzH}{z}$
Triangle AxCzH is similar to ABC This results in $\frac{AxCz}{y} = \frac{CzH}{z}$
Adding the 2 above and replacing $AxBz$ with $y-q$ and $AyBz$ with $x-q$ we end up with
$\frac{x-q}{x} + \frac{y-q}{y} = \frac{BzH}{z} + \frac{CzH}{z}$
This is the same as
$1- \frac{q}{x} + 1 - \frac{q}{y} = \frac{BzH + CzH}{z}$
But $BzH +CzH = q$
So $1- \frac{q}{x} + 1 - \frac{q}{y} = \frac{q}{z}$
Resolving the equation for unknown $q$ we get
$q = \frac{2*x*y*z}{x*y + x*z + y*z}$

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