# Triangle divided...

• Nov 8th 2010, 07:06 PM
MATNTRNG
Triangle divided...
A vertical line divides the triangle whose vertices are at (0,0), (1,1), and (9,1) into two parts; a triangle and a quadrilateral. If the areas of the triangle and the quadrilateral are equal, what is the equation of the vertical line?
• Nov 8th 2010, 08:13 PM
pickslides
for vertical line $\displaystyle x=a$ make the areas of the shapes the same.

You can try to solve $\displaystyle \displaystyle \int_a^9 f(x)~dx = \int_0^a g(x)~dx$

All you have to do is determine $\displaystyle f(x), g(x)$ which are linear combinations of the lines joining the triangle.

Can you find the equations of the 3 lines that form the triangle?

Spoiler:

$\displaystyle \displaystyle y=x,y=1,y=\frac{x}{9}$

• Nov 8th 2010, 08:47 PM
graphical approach
from the diagram,

$\displaystyle \frac{d}{1}=\frac{(9-a)}{9}$
$\displaystyle d=\frac{(9-a)}{9}$

considering the area of the small triangle,find the area of it

If A is the area of the larger triangle,

$\displaystyle A=(A+A') - A'$

$\displaystyle A1 = \frac{A}{2}$

substitute A, A1, (A+A') and A' with d and a

this is when vertical line meets AB
do the same thing when the vertical line meets CA
• Nov 9th 2010, 10:59 AM
Wilmer
Quote:

Originally Posted by MATNTRNG
A vertical line divides the triangle whose vertices are at (0,0), (1,1), and (9,1) into two parts; a triangle and a quadrilateral. If areas of the triangle and quadrilateral are equal, what is the equation of the vertical line?

Ok, now that you've been shown how, try this similar one:

A vertical line divides the triangle whose vertices are at (0,0), (7,24), and (32,24) into two parts;
a triangle and a quadrilateral. If the areas of the triangle and the quadrilateral are equal,
what is the LENGTH of the vertical line?

You'll find that all side lengths (plus the areas) are integers;
smallest case, I believe, for an all-integer.

SOLUTION (since no response!):
A(0,0), B(7,24), C(32,24)
u=7,v=32,w=24 : so A(0,0), B(u,w), C(v,w)
Vertical line length = wSQRT[(v - u) / (2v)] = 15

Smallest all-integer case is smaller than I thought: A(0,0), B(4,15), C(36,15) ; vertical line = 10
• Nov 9th 2010, 12:11 PM
pickslides
Might be easier to just find the area of the full triangle

$\displaystyle \displaystyle A = \frac{1}{2}\times 8\times \sqrt{2}\times \sin {135}\approx 0.4998$

Take half of that and solve for the smaller triangle section.

$\displaystyle \displaystyle \int_a^9 1-\frac{x}{9}~dx = \frac{0.4998}{2}$
• Nov 9th 2010, 05:40 PM
Wilmer
Quote:

Originally Posted by pickslides
Might be easier to just find the area of the full triangle
$\displaystyle \displaystyle A = \frac{1}{2}\times 8\times \sqrt{2}\times \sin {135}\approx 0.4998$

Pickslides, that should be 4 exactly, not .4998.

Easier if we let point (1,1) = (u,w) and point (9,1) = (v,w);
then A = w(v - u) / 2 = 1(9 - 1) / 2 = 4 ; agree?