Dividing a triangle in 2.

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David from Omaha

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A piece of land , triangular in shape (154 metres each side) is to be divided into 2 pieces, 1 of which is a triangle. When starting the division process , a line should be cut that starts and ends an exact number of metres from a corner. The Rom says the answer is 98 and 121 metres. I don't have a clue how to get started ! Thanks for any assistance.
 
David from Omaha said:
A piece of land , triangular in shape (154 metres each side) is to be divided into 2 pieces, 1 of which is a triangle. When starting the division process , a line should be cut that starts and ends an exact number of metres from a corner. The Rom says the answer is 98 and 121 metres. I don't have a clue how to get started ! Thanks for any assistance.
Click to expand...

Answer to what? What is the question??
 
Hello, David from Omaha!

A piece of land, triangular in shape (154 metres each side) is to be divided into
2 equal pieces, one of which is a triangle. .When starting the division process,
a line should be cut that starts and ends an exact number of metres from a corner.
The answer is 98 and 121 metres.
Click to expand...

Code: 
              C
              *
             * *
            *   *
         E *     *
          *       * 154
       y *  *      *
        *     *     *
       *        *    *
    A * * * * * * * * * B
            x      D
We have equilateral triangle \(\displaystyle ABC\) with side 154.

We want to locate point \(\displaystyle D\) on \(\displaystyle AB\) and point \(\displaystyle E\) on \(\displaystyle AC\)
. . so that the area of \(\displaystyle \Delta ADE\) is half the area of \(\displaystyle \Delta ABC,\)
. . and \(\displaystyle x = AD,\;y = AE\) are integers.
Note that: .\(\displaystyle 0 \:< \:y\:<\:154\)


The area of \(\displaystyle \Delta ABC\) is: .\(\displaystyle \frac{\sqrt{3}}{4}154^2 \:=\:5929\sqrt{3}\)

The area of \(\displaystyle \Delta ADE\) is: .\(\displaystyle \frac{1}{2}xy\sin60^o \:=\:\frac{\sqrt{3}}{4}xy\)

Hence: .\(\displaystyle \frac{\sqrt{3}}{4}xy \:=\:\frac{1}{2}\left(5929\sqrt{3}\right) \quad\Rightarrow\quad xy \:=\:11,\!858\)


We list the factors of \(\displaystyle 11,\!858\!:\: (1,11,\!858),\;(2,5929),\;(7,1694)\;\cdots\)

The only pair in the domain is: .\(\displaystyle (x,y) \:=\: (98,121)\)
 
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David from Omaha said:
A piece of land , triangular in shape (154 metres each side)
is to be divided into 2 pieces, There is nothing in this statement about the pieces
being equal in area.


1 of which is a triangle.


When starting the division process ,


a line should be cut that starts and ends an exact number of metres from a corner.

1) An "exact number of meters" should mean that it is a value that is not from
a measurement. For instance, 1 inch is defined as exactly 2.54 cm. So, I don't
see a necessary requirement for the line segments on each of two sides of the
original triangle, ** respectively, to have an integer number of units.


2) "a line should be cut that starts and ends . . . from a corner"

A "corner" would be a vertex of the triangle. For instance, soroban's offer
of a solution does not have the dividing line segment coming out of a "corner."


The Rom says the answer is 98 and 121 metres.
Click to expand...

I call foul on the presentation of this problem with its ambiguous descriptions
and assumed conditions.


** This is where the two line segments on the original triangle are
created by where the dividing line intersects with each of two sides, respectively,
of the original triangle.
 
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A line from one vertex of a triangle to the opposite side, that cuts the triangle into two equal areas, must go through the midpoint of that side. That is because the area of a triangle is "(1/2) base times height". Since the two triangles have the same height (perpendicular distance from the side to the vertex) the two bases must have the same length.
 
HallsofIvy said:
A line from one vertex of a triangle to the opposite side, that cuts the triangle into two equal areas, must go through the midpoint of that side. That is because the area of a triangle is "(1/2) base times height". Since the two triangles have the same height (perpendicular distance from the side to the vertex) the two bases must have the same length.
Click to expand...

Thanks for a brilliant solution. No one else could do it ! Smart asses ! Cricket ??????? A girls version of baseball. Is cricket in the olympic games ?
 
Hello, HallsofIvy!


I knew that any median divides the triangle into two equal pieces
. . where both pieces are triangles.

But please note the bold-face one in the statement of the problem
. . in my post.
 
Denis said:
Ya ok, but why mention that?
The OP's problem has nothing to do with that:
we have a line creating a triangle and a quadrilateral of equal area.
No ??????
Click to expand...

No. See my post (#5).

And another point to (re)emphasize, certain allegedly intended aspects of the
problem are inconsistent with each other. Even if the problem meant to state
that there would be exactly one triangle and some other differently-sided
polygon created by the dividing line, it's not going to happen with a line that
comes out of a corner (read: vertex).
That line "emanating from the corner" would make exactly two triangles.



And this "equal area" myth keeps getting propagated.

Source (from the original post):

"is to be divided into 2 pieces"

Does this phrase mean that the pieces are to be necessarily the same area?

Nope.
 
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Denis said:
A piece of land , triangular in shape (154 metres each side) is to be divided into 2 pieces,
1 of which is a triangle. When starting the division process , a line should be cut that starts
and ends an exact number of metres from a corner. The Rom says the answer is 98 and 121 metres.
I don't have a clue how to get started ! Thanks for any assistance.
​
That's the original post: I don't see "vertex" in there...
That's correct! You don't see it, and it's not there.
A corner of a triangle is a vertex. So! If someone started
to be liberal with "corner," they might argue that it would
be a point on the triangle that is just shy of the midpoint
of the side it would emanate from. What a slippery slope
(no pun intended)!
Hm? Where would the corner of this triangle begin, and
where in its interior or on its perimeter would it end?

IF the line started from a vertex, then this would be a no-brainer
since we're dealing with an equilateral triangle.

There is no no-brainer about that, as there
is still is nothing about dividing into 2 equal areas. It would potentially be
a line from a vertex to somewhere to the opposite side. And the part of it
being divided into one triangle wouldn't preclude the other part from being
a triangle, as it didn't have an equivalent phrasing
such as "exactly one triangle."


If we're to guess at the OP's meaning, then (to me) Soroban's
way is the only correct way. Guessing at the OP's meaning isn't an option here.
The problem statement is too flawed.
"If we're to guess...." And if wishes were fishes. No free passes. No hand-waving
to move on ahead.

That's it from me...ain't losing any more time on this...
Fine, I'm not leaving, and I'm not "wasting" time on this.
Click to expand...



Original poster, David from Omaha,

is your question verbatim (aside from spelling out or not of numbers or units)?

If you presented to us accurately what you were given,
then the problem should be tossed out.
 
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