BMO 1993 Q2

 Q2. A square ABCD of side 1 is divided into two parts by the diagonal AC.

Find the length of the straight line, of minimum length,  that divides triangle ABC into two parts of equal area.

The first thing to do with this problem is to do a sketch and introduce some variables to characterise the approach to a solution. See the sketch below:

Diagram showing the construction and equation corresponding to the specified division of area.

Here I introduce variables \(a\) and \(b\) then the condition that the line cuts the area of the triangle into two equal area pieces gives:

\[A=\frac{a^2}{2}+a(1-a)+\frac{(b-a)(1-a)}{2}=\frac{1}{4}\]

Which may be rearranged to give \(b\) in terms of \(a\):

\[b=\frac{2a-1}{2a-2}\]

The length of the cut \({l}\) satisfies:

\[l^2=(b-a)^2+(1-a)^2\]

Then is we find \(a\) that miinimises \(l^2\) it also minimises \(l\), and this occurs when \(\frac{dl^2}{da}=0\).

After a lot of tedious algebra this gives:

\[\frac{dl^2}{da}=\frac{8a^4-32a^3+48a^2-32a+7}{2a^3-6a^2+6a-2}=0\]

Nowthe denominator is \(2(a-1)^2\) which is never zero on \( (0,1) \) and so can be multiplied out to leave:

\[8a^4-32a^3+48a^2-32a+7=0\]

Now we have the trixy part ... this is a quartic, and even though there is a "formula" for the roots of a quartic (Ferrari/Cardarno method) it is impractical to use in a exam setting (and I don't remember it anyway). But look at the coefficients of the non-constant terms, these are eight times the fourth row of Pascal's triangle, so:

\[8a^4-32a^3+48a^2-32a+7=8(a-1)^4+7-8=0\]

so if \(a\) is real: \((a-1)=\pm \frac{1}{2^{3/4}}\), or \(a=1-\frac{1}{2^{3/4}}\) (as we want the root between \(0\) and \(1\) ).

Then \(b=1-\frac{1}{2^{1/4}}\), and \(l=\sqrt{\sqrt{2}-1}\approx 0.6436\).