Thread: Pythagorean theorem and an equilateral triangle

Let ABC be an equilateral triangle and P be a point inside this triangle such that PA=x, PB=y and PC=z.

If , find the length of the sides of triangle ABC in terms of x and y.

How do we go about this? AFAIK, the Pythagorean theorem can be applied to Right Angle Triangles. I am sure there is a Right Angle Triangle hidden somewhere in there...

1) It is an equilateral triangle. The Pythagorean Theorem will not do you much direct good.

2) If you construct a perpendicular on each of the three sides, through point P, you should start seeing Right Triangles in your dreams.

Yes I know I have to do something like that. But where in heavens name should point P be?

Originally Posted by cosmonavt

Let ABC be an equilateral triangle and P be a point inside this triangle
such that PA=x, PB=y and PC=z.
If , find the length of the sides of triangle ABC in terms of x and y.

Say you take those 3 : PA, PB and PC, and form a triangle with them, what will you get?
A right triangle, agree? If not, then all I can say is you need basic help from your teacher.

OK, do some work:
draw equilateral triangle ABC, insert point P inside so that AP=3, CP=4 and BP = 5; like:

Code:

                 A
 
 
               x=3
 
 
              P
 
 
       y=4            z=5
 
 
 
C                                  B

x^2 + y^2 = z^2
3^2 + 4^2 = 5^2
9 + 16 = 25
25 = 25 ; got that?

Now go calculate the side length of triangle ABC.
Once you "see" what's going on, do over in terms of x,y,z.

Well, before you edited ur post, I actually did some research on it.

As a general rule, for every equilateral triangle of side , the area is given by

If we make a point P at any random location in the triangle and draw three lines through it each perpendicular to a side, we end up in something like this (ignore the side lengths 1. Just assume them to be )



Now the area of each separate triangle formed is

x a x h1

x a x h2 and

x a x h3

respectively.

Hence the equation:
= x a x h1 + x a x h2 + x a x h3

Multiplying both sides by :

= h1 + h2 + h3

in our case...

= x + y + z

and the second equation is:



Now we can solve the two simultaneously to obtain a.

BUT

there are infinite values which satisfy the equation

Putting those values in the equation = x + y + z yields infinite values for a also.

Originally Posted by cosmonavt

> Well, before you edited ur post, I actually did some research on it.

The edit was to ADD stuff; all I had before the edit is still there...

> If we make a point P at any random location in the triangle......
>there are infinite values which satisfy the equation z^2 = x^2 + y^2

Yes; BUT in your problem's case, P cannot be at random,
since z^2 = x^2 + y^2; so x,y,z MUST be sides of a right triangle.

> Putting those values in the equation = x + y + z yields infinite values for a also.

So what? You get a in terms of x,y,z; that's what you have been asked to do...

To wrap up, the problem is worded:
"find the length of the sides of triangle ABC in terms of x and y."
(notice that "z" is not part of final expression)

a = side length, k = SQRT(3)/2

a = [x + y + SQRT(x^2 + y^2)] / k

Yeah I just realized it :P Thanks man! I totally forgot the "in terms of x and y" part.

You're doing pretty good...what grade are you in?

Ahahaha, I am an A level student. U can say grade 12 in my country. But we are not taught much geometry in here. 90% of our syllabus is calculus. So yeah, it is something to be pleased at.

Originally Posted by cosmonavt

Ahahaha, I am an A level student. U can say grade 12 in my country. But we are not taught much geometry in here. 90% of our syllabus is calculus. So yeah, it is something to be pleased at.

"U" is not a word; should be "You":
may be important if you apply for a job in English !!

Originally Posted by cosmonavt

If we make a point P at any random location in the triangle

Another important word to get right.

You mean "an arbitrary point", not "a point...at any random location".

"Random" and "arbitrary" often are confused, but they are necessarily different.