How to bisect an isosceles trapezoid into two equal area parts

Question

Given an isosceles trapezoid:

I want to draw a line parallel to the bases (that is, parallel to and in between AD and BC) such that the top half and the bottom half both have equal area.

Specifically, if we define the height of the trapezoid as 1, I want to know how far from the longer base this bisector will fall. Let's call this value m.

For an isosceles trapezoid where the shorter base is some factor k shorter than the other, I suspect that m will be a function only of k.

Here's a more detailed diagram:

Here are some (possibly incorrect) equations describing this:

• Area of ABCD = ((a + ka) / 2) * 1
• Area of EBCF = ((a + ja) / 2) * m
• Area of AEFD = ((ja + ka) / 2) * (1 - m)
• Area of EBCF = Area of AEFD
• Area of EBCF + Area of AEFD = Area of ABCD

Intuitively, I would expect j and a to cancel out of these equations, and to be able to get m solely in terms of k. But I'm having trouble figuring out how to get there.

Answer 1

The important thing is to note that $(a-ka)=\frac {ja-ka}{1-m}$. You can prove that by dropping perpendiculars from $A$ and $D$ to $BC$ and observing the triangles on each end are similar. You can take $a$ out of this to get $1-k=(1-m)(j-k)$ Couple this with $(1+j)m=(j+k)(1-m)$ from the equal areas and you have two equations in three unknowns, which will get you the one parameter solution set you are looking for. $$1-k=\frac{j-k}{1-m}\\(1+j)m=(j+k)(1-m)\\ 1-m-k+km=j-k\\m+jm=j+k-jm-km\\1-m+km=j\\2jm-j=k-km-m\\(2m-1)(1-m+km)=k-km-m$$which gives you a quadratic in $m$

Answer 2

Hint: Try breaking the trapezoid up into a rectangle and an isosceles triangle. I think that you will find that as the height increases, the rectangle's area will increase linearly, and that the triangle's area will increase quadraticly.

Answer 3

This builds on Ross Milkman's answer and makes it more explicit. I suggested it as an edit to his, but the edit was rejected.

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The important thing is to note that $(a-ka)=\frac {ja-ka}{1-m}$. You can prove that by dropping perpendiculars from $A$ and $D$ to $BC$ and observing the triangles on each end are similar. You can take $a$ out of this to get $1-k=\frac {j-k}{1-m}$:

$$1-k=\frac{j-k}{1-m}\\ 1-m-k+km=j-k\\1-m+km=j$$

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Couple this with $(1+j)m=(j+k)(1-m)$ from the equal areas and you have two equations in three unknowns, which will get you the one parameter solution set you are looking for.

$$\\(1+j)m=(j+k)(1-m)\\m+jm=j+k-jm-km\\2jm-j=k-km-m\\j(2m - 1)=k-km-m$$

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And then, substitute the first into the second to get a quadratic in $m$:

$$\\(2m-1)(1-m+km)=k-km-m\\2m-1-2m^2+m+2km^2-km=k-km-m\\(2k-2)m^2+4m-(1+k)=0$$

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Then, use the quadratic equation to get $m$ in terms of $k$:

$$\\m=\frac{-4±\sqrt{16-4(2k-2)(-1-k)}}{2(2k - 2)}$$

So then, for $k=0.1$: $m=.321518311$ or $m=1.900703911$

Since $m$ can't be a value larger than the height of the trapezoid itself, the first answer is the only one that has meaning in this context.

Answer 4

Generalizing to arbitrary trapezoids doesn't make things any more difficult, so we won't assume the trapezoid is isosceles or that it has height $1$. The key fact is that $$ (ja)^2=\frac{a^2+(ka)^2}{2}. $$ This can be seen by extending the lines $BA$ and $CD$ until they meet. (The case where the sides are parallel is trivial.) Call the meeting point $O$. Then triangles $BOC$, $EOF$, and $AOD$ are similar, which implies that their areas are equal to some proportionality constant multiplied by $a^2$, $(ja)^2$, and $(ka)^2$. The areas of trapezoids $BCFE$ and $EFDA$ are therefore proportional to $a^2-(ja)^2$ and $(ja)^2-(ka)^2$ with the same proportionality constant. Setting these equal gives the key fact. As a result, $$ j=\sqrt{\frac{1+k^2}{2}}. $$ To find $m$, use that the reduction in horizontal width is proportional to the height to get $$ \frac{m}{h}=\frac{a-ja}{a-ka}=\frac{1-j}{1-k}, $$ with the result $$ m=\frac{1-\sqrt{\frac{1+k^2}{2}}}{1-k}h. $$

Historical aside: The key fact was known in Old Babylonian times (~2000 BCE– ~1600 BCE). As an example, it was used in the breathtakingly elegant solution to the problem on cuneiform tablet VAT 8512, which is explained in Jens Høyrup's book Algebra in Cuneiform: Introduction to an Old Babylonian Geometrical Technique. Problems given in Old Babylonian scribal education were generally contrived so as to have exact, finite representations in base-60 notation. Two whole-number "trapezoidal triples" $(a,ja,ka)$ are $(7,5,1)$ and $(17,13,7)$. The triple $(51,39,21)$ that appears on VAT 8512 is a multiple of the latter. If reconstructions of the damaged number at the top of the tablet are to be believed, IM 58045 from the Old Akkadian period (2400 BCE–2250 BCE) may provide an even older example of this triple and is, in fact, one of the oldest known mathematical tablets. Finally, note that trapezoidal triples are close cousins of Pythagorean triples: if $p^2+q^2=r^2$, then $(q+p)^2+(q-p)^2=2r^2$ so that the Pythagorean triple $(p,q,r)$ corresponds to the trapezoidal triple $(q+p,r,q-p)$. So $(3,4,5)$ corresponds to $(7,5,1)$ and $(5,12,13)$ corresponds to $(17,13,7)$.