# CTK Insights

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21 May

### Area and Perimeter Splitters in a Triangle

I have recently posted a constructive solution by Dan Shved to the problem of finding a line through a given point ($D)$ that halves the area of a given triangle $(\Delta ABC).$

There is a calculus based proof of existence (see, e.g., Honsberger.) For $D$ outside $\Delta ABC$, there are two rays emanating from D that form the smallest angle containing the triangle. The rays from $D$ in this angle cross and divide the triangle into two parts, say, left and right (looking from $D).$ If $\alpha$ is the angle form one of the extreme rays that describes the position of a ray through D, then the difference in the "left" and "right" areas is a function $f(\alpha)$ of $\alpha.$ The function is continuous and is negative at one extreme location and is positive at the other one. By the Bolzano Theorem, $f$ vanishes somewhere between the two extremes. For this angle, i.e., for this position of the ray, the left and right areas cut off by the ray from the triangle are equal - and this is exactly what is needed.

For point $D$ within the triangle, the proof changes only slightly. Lines through $D$ divide the triangle into "left" and "right" parts. By choosing one of the lines and rotating it $180^{\circ}$ degrees either direction, we get the same line but with the "left" and "right" parts interchange such that the difference in the two areas flips the sign, implying - again by the Bolzano Theorem - that somewhere along the way the two areas were equal.

There is of course a huge conceptual difference between knowing a problem has a solution and being able to actually produce that solution. However the generality of the foregoing proof has its advantages. For example, it is clear that - slightly modified - it will solve the problem of establishing the existence of a perimeter splitter through a given point.

It is then meaningful to ask whether a line could be simultaneously an area and a perimeter splitter, and seek a characterization of such lines if they exist. As a matter of fact, they do, and there may be $1,$ $2,$ or $3$ such lines. This was shown by A. Todd a senior at university. In truth, the problem has an interesting history.

In 1994, A. Shen published in The Mathematical Intelligencer a list of the so-called coffin problems - elementary looking problems with hard solutions - that were offered to Jewish applicants at the entrance exams at the Mathematics Department of the Moscow State University in the 1970s and 1980s. Problem 5 in Shen's list requires to Draw a straight line that halves the area and perimeter of a triangle. I. Vardi published solutions to those problems in a postscript file. (Unfortunately, my attempts to install a postscript reader led to my computer infestation that much diminished my interest in Vardi's solutions.) However, this was included as the first chapter in M. Shifman's book now available online. The solution there is algebraic and indeed lengthy. It is constructible since all the formulas that appear in the solution are quadratic.

It is not hard to establish that any line that halves both the perimeter and the area of a triangle has to pass through its incenter, see, for example, [Honsberger, Kodokostas, Kung, Todd]. The proof below is based on the area splitter construction. It makes use (like probably all other proofs) of the fact that the area $S$ of a triangle can be found from $S=rs,$ where $r$ is the inradius, $s$ the semiperimeter of the triangle.

Assume that $FG$ is the area splitter. Then $r(BF+BG)/2$ is the sum of the areas of triangles $BIF$ and $BIG$ which differs from the area of $\Delta BFG$ unless the incenter $I$ lies on $FG.$

It is as easily established that passing through the incenter is a necessary and sufficient condition of a perimeter splitter to be an area splitter, and the statements are reversible.

The following diagram shows Dan Shved's solution adapted to the case where $D=I$.

The diagram depicts the configuration where two solutions $FG$ and $F'G'$ cross sides $AB$ and $BC$. These are symmetric in the bisector of angle at $B.$

In a thorough investigation, D. Kodokostas tabulated all the possible combinations of the angles of $\Delta ABC$, proving there are cases when the number of solutions is exactly $1,$ $2$, or $3.$

### References

1. M. Shifman (ed), You failed your math test, Comrade Einstein, World Scientific, 2005
2. R. Honsberger, Mathematical Delights, MAA, 2004, 71-74
3. D. Kodokostas, Triangle Equalizers, Mathematics Magazine 83, No. 2 (April, 2010) 141–146; available at http://dx.doi.org/10.4169/002557010X482916
4. S. Kung, Proof Without Words: A Line through the Incenter of a Triangle, Mathematics Magazine 75, No. 3 (Jun., 2002) 214
5. A. Shen, Entrance Examinations to the Mekh-mat, The Mathematical Intelligencer, Vol. 16, No.4 (1994) 6–10
6. A. Todd, Bisecting a Triangle, Pi Mu Epsilon Journal, Vol. 11, No. 1 (1999) 31–37
7. A. Todd, New and Letters, Mathematics Magazine 84 (2011) 396
8. I. Vardi, Mekh-mat entrance examination problems

× three = 3