Love Equilateral Triangles?

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Love Equilateral Triangles?

by dtweah » Sat May 09, 2009 7:10 am
Let A,B,C be vertices of an equilateral triangle with sides of length 1, and P a point interior to the triangle. If a,b,c are the distances from P to the sides of the triangle then a + b + c equals:

(a) sqrt(3)/2
(b) 3/4
(c) sqrt(2)
(d) 2/3
(e) Depends on the location of P

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by DanaJ » Sat May 09, 2009 8:25 am
Intriguing... Yet fun, since it's geometry. All you need to do is draw the picture, it's smooth sailing after that one.

As you can see, P is randomly placed. The distances from P to the sides are actually the lengths of perpendicular segments (since that's the definition of the distance from a point to a line: the perpendicular segment drawn from that point to the line).

Now, notice how P breaks your equilateral triangle into three smaller triangles that share P as vertex: APB, APC, BPC. Again, notice that a is the height of triangle BPC, b is the height of triangle APC and c is the height of triangle APB.

What does this mean? Just look at the areas for the three smaller triangles. Their areas will be:

For APB: c*AB/2 or c/2 (since AB is the side of the triangle and is 1).

For APC: b*AC/2 or b/2

For BPC: a*BC/2 or a/2.

As you can plainly see from the figure below, when you add up the areas of your three smaller triangles, you get the big equilateral triangle. The area of this one will be sqrt(3)/4 (following the standard formula for an equilateral triangle, i.e. area = (side)^2*sqrt(3)/2).

This means that a/2 + b/2 + c/2 = sqrt(3)/4 ----- multiply both sides by 2 to get what you're looking for:

a + b + c = sqrt(3)/2 - with answer A.




Side note: I actually solved this in about 30 seconds because I knew a few shortcuts.
It's one of the basic rules of geometry that the sum of distances from interior point P to the sides of a triangle is constant, no matter the type of triangle (equilateral, right etc....). Here in sunny Romania, they teach you this rule in 6th grade. However, it's not one of the popular theorems, so most people forget about it. I remembered it because I had a particularly sadistic maths teacher....

Anyway, once you know this, you can apply another rule. In any triangle, the centroid of a triangle (the point where its medians intersect) has a "special" little place. As wikipedia tells us, two-thirds of the length of each median is between the vertex and the centroid, while one-third is between the centroid and the midpoint of the opposite side.

Since the centroid is a point inside the triangle, the sum of distances from it to the sides of the triangle is exactly what you're looking for. Here comes the equilateral triangle part: in such a triangle, the median and height coincide, and most of us know that the height of an equilateral triangle is [(side)*sqrt(3)/2]. This means that the distance between the centroid and the three sides will always be one third of the height (or median), i.e. (side)*sqrt(3)/6. Since you've got three (equal) such segments, a + b + c in our problem will be 3*(side)*sqrt(3)/6 = (side)*sqrt(3)/2.

I sure hope someone understands this...


EDIT: I added the image in my other post below. Thanks Eric!
Last edited by DanaJ on Sun May 10, 2009 12:02 am, edited 1 time in total.

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by DanaJ » Sat May 09, 2009 8:28 am
I don't know why, but it won't accept my file. It's a JPEG with 5.95 KB and I don't believe that I've uploaded too much...

If anyone wants to see the picture, PM with email address and I'll answer as quickly as possible.

Image