Point in triangle...

• Jul 23rd 2010, 04:07 AM
Wilmer
Point in triangle...
Triangle ABC with BC=10, AC=11 and AB = 12.

Point P is inside triangle ABC, such that triangles APB, APC and BPC have equal areas.

What are lengths od AP, BP abd CP?

Work so far:
area of ABC = 33SQRT(39) / 4
triangle APB : height from P = (33SQRT(39)/2) / 36
triangle APC : height from P = (33SQRT(39)/2) / 33
triangle BPC : height from P = (33SQRT(39)/2) / 30

From that, how would you calculate lengths of AP, BP and CP?
I keep going around in circles! Thanks.
• Jul 23rd 2010, 04:14 AM
undefined
Quote:

Originally Posted by Wilmer
Triangle ABC with BC=10, AC=11 and AB = 12.

Point P is inside triangle ABC, such that triangles APB, APC and BPC have equal areas.

What are lengths od AP, BP abd CP?

Work so far:
area of ABC = 33SQRT(39) / 4
triangle APB : height from P = (33SQRT(39)/2) / 36
triangle APC : height from P = (33SQRT(39)/2) / 33
triangle BPC : height from P = (33SQRT(39)/2) / 30

From that, how would you calculate lengths of AP, BP and CP?
I keep going around in circles! Thanks.

Maybe this will help

Median (geometry) - Wikipedia, the free encyclopedia (subheading Equal-area division)
• Jul 23rd 2010, 04:27 AM
Quote:

Originally Posted by Wilmer
Triangle ABC with BC=10, AC=11 and AB = 12.

Point P is inside triangle ABC, such that triangles APB, APC and BPC have equal areas.

What are lengths od AP, BP abd CP?

Work so far:
area of ABC = 33SQRT(39) / 4
triangle APB : height from P = (33SQRT(39)/2) / 36
triangle APC : height from P = (33SQRT(39)/2) / 33
triangle BPC : height from P = (33SQRT(39)/2) / 30

From that, how would you calculate lengths of AP, BP and CP?
I keep going around in circles! Thanks.

Alternatively, you could play with co-ordinate geometry.
Placing B at the origin, C is at (10,0).
Then locate the co-ordinates of A.
You've worked out the triangle areas.
You've got the height of P from the line [BC].
Then P lies on the line y=h.
The intersection of the line parallel to [AB] with appropriate perpendicular distance from [AB], with y=h,
or the intersection of the line parallel to [AC] with y=h, correct distance from [AC]
locates P.
• Jul 24th 2010, 10:37 AM
Wilmer
Had a closer look at this; turns out easy enough.

We have triangle ABC with BC = a =10, AC = b = 11 and AB = c = 12 ; point P inside.

STEP#1:
Make B(0,0) and A(12,0) ; and C(g,h)
Get the area k.
h = 2k / c
g = SQRT(a^2 - h^2)

STEP#2:
Make P(x,y)
x = (c + g) / 3
y = h / 3
NOTE: since 3 inside triangles equal in area, then P is the centroid!

STEP#3:
AP = SQRT((c - x)^2 + y^2) ; ~6.9121
BP = SQRT(x^2 + y^2) ; ~6.3857
CP = SQRT((x - g)^2 + (h - y)^2) ; ~5.7542

Thanks for the hints fellows.
• Jul 26th 2010, 03:34 AM
simplependulum
Note that $P$ is the centroid of the triangle , to calculate $AP , BP ,CP$ . We can find the length of the medians first , then obtain the answers by multiplying $\frac{2}{3}$ .

If we let $M$ be the mid-point of $AB$

By Apollonius' Theorem , $AC^2 + BC^2 = 2( AM^2 + CM^2)$

or $CM^2 = \frac{a^2 + b^2 }{2} - \frac{c^2}{4}$

Then substitute the values of $a,b,c$ ...
• Jul 26th 2010, 06:17 AM
Wilmer