Equally Trapped

Relevant wiki: Trapezoid

Suppose that it is possible. Start with a trapezoid, and let the parallel sides of the trapezoid have lengths $b_1$ and $b_2,$ and fix the height of the trapezoid at $1.$ (If the point can be found for a trapezoid with height $1,$ then it can be found for any similar trapezoid.)

The area of the trapezoid is $\frac{1}{2}(b_1+b_2).$ This means that each of the triangles has area $\frac{1}{8}(b_1+b_2).$

Let $y$ be the height of the triangle with base $b_2.$ This triangle has area $\frac{1}{2}b_2y.$ Now, the triangle with base $b_1$ has height $1-y.$ The area of this triangle is $\frac{1}{2}b_1(1-y).$ These two triangles have equal areas:

$\begin{aligned} \frac{1}{2}b_2y &= \frac{1}{2}b_1(1-y) \\ y &= \frac{b_1}{b_1+b_2} \end{aligned}$

So the area of each triangle can also be expressed as $\frac{b_1b_2}{2(b_1+b_2)}.$ Now we have:

$\begin{aligned} \frac{1}{8}(b_1+b_2) &= \frac{b_1b_2}{2(b_1+b_2)} \\ (b_1+b_2)^2 &= 4b_1b_2 \\ b_1^2-2b_1b_2+b_2^2 &= 0 \\ (b_1-b_2)^2 &= 0 \\ b_1 &= b_2 \end{aligned}$

However, if the two bases are equal and parallel, then this causes the shape to become a parallelogram. This contradicts the assertion that the shape is a trapezoid. Therefore, it is not possible.

Relevant wiki: Area Of Triangle

Label the trapezoid ABCD so that BC and AD are the parallel sides and BC<AD Let the inside point be P and the vertical height of the trapezoid be h.

Let the length of BC be a

AD is some multiple of BC so AD=Ka

The only way triangles BCP and ADP can have the same area is if

vertical height of triangle BCP=Kh/(K+1)

So area of triangle BCP= area of triangle ADP

= aKh/2(k+1)

Now area of trapezoid = a(K+1)h/2

One quarter of this area = a(K+1)h/8

So the question is now can

aKh/2(K+1) = a(K+1)h/8

This simplifies to the quadratic equation

K^2-2K+1 = 0

Solution is K = 1

So the areas can only be equal if K=1

ie if BC = AD

so the shape is a parallelogram and not a trapezoid

So it can never be true

There may be a very complicated and accurate way to find the answer, but there is a simpler way. The question says that there is only one pair of parallel sides. If there is only one pair, how could four equal triangle result from splitting the trapezoid

Let ABCD be the trapezoid and let AB, CD be the pair of parallel edges, and let M be this supposed midpoint.

If the triangles ABM and CDM are to have equal area, then M may be any point on a particular line parallel to both AB and CD and which is closer to the longer of the two (the exact ratio is irrelevant)

Now consider the shape ABMD. If this is a quadrilateral then it must be a parallelogram if ABM and AMD are to have equal areas. This is not allowed as CDM would then have no area.

Then it must be a triangle and the same will apply for ABCM. Therefore M must be the point of intersection of the diagonals AC and BD.

But then ABM and CDM are then similar, and if they are to have equal areas then they must be congruent. Therefore AB must be equal to CD which is not allowed if ABCD is to be a trapezoid.

Let the upper and lower sides and altitude of the trapezoid be a, b, and h. AlsoLet the altitudes of upper and lower triangle be $h_1$ , and $h_2$ . Therefore, a $h_1$ /2 = b $h_2$ /2= (a + b)h/8, 4a $h_1$ = 4b $h_2$ = (a + b)h, $h_1$ + $h_2$ = h, $h_1$ = h – $h_2$ , so -4a $h_2$ + 4ah = (a + b)h, so $h_2$ = (3a – b)h/(4a) = (a + b)h/(4b). Then 3ab – $b^2$ = $a^2$ + ab, or $a^2$ – 2ab + $b^2$ = 0, or $(a - b)^2$ = 0. Therefore a = b, whichis impossible for a trapezoid.