Triangle divided...

**MATNTRNG** · November 8th 2010, 07:06 PM

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?

**pickslides** · November 8th 2010, 08:13 PM

for vertical line $x=a$ make the areas of the shapes the same.

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

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

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

Spoiler:

**BAdhi** · November 8th 2010, 08:47 PM

from the diagram,

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

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

If A is the area of the larger triangle,

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

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

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

**Wilmer** · November 9th 2010, 10:59 AM

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

**pickslides** · November 9th 2010, 12:11 PM

Might be easier to just find the area of the full triangle

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

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

$\displaystyle \int_a^9 1-\frac{x}{9}~dx = \frac{0.4998}{2}$

**Wilmer** · November 9th 2010, 05:40 PM

Originally Posted by pickslides

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

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?

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