A cute triangle problem

by nsato, Nov 20, 2010, 12:18 AM

Let $P$ be a point inside equilateral triangle $ABC$ . Let $A_1$ , $B_1$ , and $C_1$ be the feet of the altitudes from $P$ to sides $BC$ , $AC$ , and $AB$ , respectively.

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(a) Show that the sum of the perimeters of triangles $AC_1 P$ , $BA_1 P$ , and $CB_1 P$ is equal to the sum of the perimeters of triangles $AB_1 P$ , $CA_1 P$ , and $BC_1 P$ .

(b) Show that the sum of the areas of triangles $AC_1 P$ , $BA_1 P$ , and $CB_1 P$ is equal to the sum of the areas of triangles $AB_1 P$ , $CA_1 P$ , and $BC_1 P$ .

I'll post my solution in a few days.

[jump]Learn more after the jump![/jump]

Both parts follow immediately from drawing lines through $P$ that are parallel to the sides.

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This post has been edited 1 time. Last edited by nsato, Dec 7, 2010, 7:01 PM

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Not a beast construction, but oh well...

Let $X=PB_1\cap CA_1$ . Then by simple angle chasing, $PX=2PA_1$ and
$CB_1=\frac{PX+PB_1}{\sqrt3}=\frac{2PA_1+PB_1}{\sqrt3}.$ Similar expressions can be found for $AB_1,AC_1,BA_1,BC_1,CA_1$ (note that $PA_1,PB_1,PC_1$ uniquely determine everything.)

This post has been edited 6 times. Last edited by math154, Nov 24, 2010, 10:26 PM

by math154, Nov 24, 2010, 9:11 PM

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Draw lines through P parallel to the sides.

We want AB1+BC1+CA1=AC1+BA1+CB1 for a). We want to prove AB1-BC1+BA1-BC1+CB1-CA1=0. This is easy to see, since you have a bunch of equals sides cancel each other

For part b), note the parallel lines divide the triangle into 3 triangles and 3 parallelograms. The lines of the triangles divide each of them in half. The its easy to see the sum of their areas are the same.

by abacadaea, Nov 25, 2010, 3:16 AM

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A cute triangle problem

by nsato, Nov 20, 2010, 12:18 AM

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