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The lines which bisect the area of a triangle form an envelope as shown in this picture

alt text

It is not difficult to show that the ratio of the area of the red deltoid to the area of the triangle is $$\frac{3}{4} \log_e(2) - \frac{1}{2} \approx 0.01986.$$

But this is also $$\sum_{n=1}^{\infty}\frac{1}{(4n-1)(4n)(4n+1)}.$$

Is there any connection between the series and the deltoid? Or is it just a coincidence?

(I asked this at MathOverflow almost 18 months ago and got no response)

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  • $\begingroup$ Uhm, "it's not difficult to show" .. I feel a bit embarrassed not being able to show anything. What's so easy about it? Can you give me a hint or a reference? $\endgroup$ – Han de Bruijn Apr 8 '14 at 12:26
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    $\begingroup$ @Han de Bruijn: You can use affine transformations to show the ratio is invariant. Suppose the corners of the triangle are $(0,0)$, $(0,1)$, $(1,0)$ so the reflection of this diagram. Then green area bisectors which do not cross the base are of the form $y=c+x(2(1-c)^2-1)$ for $0 \lt c \lt \frac12$ and the lower red envelope is of the form $y=1-x-\dfrac{1}{8x}$ between $(\frac14,\frac14)$ and $(\frac12,\frac14)$. You can then find the area ratio by integration as $\displaystyle 6\left(\frac{7}{96}-\int_{1/4}^{1/2}\left(1-x-\dfrac{1}{8x}\right)dx\right).$ $\endgroup$ – Henry Apr 8 '14 at 14:18
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+50
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Since $|\triangle BPQ|=\frac12\sin(B)|\overline{BP}||\overline{BQ}|$, to maintain $|\triangle BPQ|=\frac12|\triangle ABC|$, we need $|\overline{BP}||\overline{BQ}|=\tfrac12|\overline{BC}||\overline{BA}|$.

$\hspace{3.4cm}$enter image description here

In the diagram above, the endpoints of the lines are parametrized by $$ P_t=\frac{1-t}2B+\frac{1+t}2C\tag{1} $$ and $$ Q_t=\frac{t}{1+t}B+\frac1{1+t}A\tag{2} $$ for $t\in[0,1]$.

The points on each line are parametrized by $$ R_t(s)=(1-s)P_t+sQ_t\tag{3} $$ To find a point on the envelope of the family of lines parametrized by $t$ we need to find where $$ \frac{\partial}{\partial t}R_t(s)=\frac{1-s}2(C-B)+\frac{s}{(1+t)^2}(B-A)\tag{4} $$ is parallel to $$ \frac{\partial}{\partial s}R_t(s)=\frac1{1+t}(A-B)+\frac{1+t}2(B-C)\tag{5} $$ Setting the cross product to $0$, we get that $$ \frac{1-s}{2(1+t)}-\frac{s}{2(1+t)}=0\tag{6} $$ which happens when $s=\frac12$. That is, the locus of the envelope is the midpoints of the line segments $$ M_t=B+\frac{1+t}4(C-B)+\frac1{2(1+t)}(A-B)\tag{7} $$ $\hspace{3.4cm}$enter image description here

Subtracting the center of the triangle (and of the deltoid) at $M_\triangle=\frac13(A+B+C)$ yields $$ M_t-M_\triangle=\left(\frac{1+t}4-\frac13\right)(C-B)+\left(\frac1{2(1+t)}-\frac13\right)(A-B)\tag{8} $$ Thus, $3$ times the area of $\frac13$ of the deltoid is $$ \begin{align} &\frac32\int_0^1(M_t-M_\triangle)\times(M_t-M_\triangle)'\,\mathrm{d}t\\ &=3|\triangle ABC|\int_0^1\left[\frac14\left(\frac1{2(1+t)}-\frac13\right)+\frac1{2(1+t)^2}\left(\frac{1+t}4-\frac13\right)\right]\,\mathrm{d}t\\ &=\frac{3\log(2)-2}{4}|\triangle ABC|\tag{9} \end{align} $$ where $|\triangle ABC|=\frac12|(A-B)\times(C-B)|$.

Since the deltoid, as parametrized in $(7)$, is a convex combination of the vertices, it transforms with any affine linear transformation of the vertices. Thus, the ratio of areas does not depend on the particular location of the vertices.


$$ \begin{align} \sum_{k=1}^\infty\frac1{(4k-1)4k(4k+1)} &=\frac12\sum_{k=1}^\infty\left(\frac1{4k-1}-\frac1{2k}+\frac1{4k+1}\right)\\ &=\lim_{n\to\infty}\frac12\left(\color{#C00000}{\sum_{k=1}^{2n}\frac1{2k+1}-\sum_{k=1}^{2n}\frac1{2k}}+\color{#00A000}{\sum_{k=n+1}^{2n}\frac1{2k}}\right)\\ &=\frac12\left(\color{#C00000}{\log(2)-1}+\color{#00A000}{\frac12\log(2)}\right)\\[3pt] &=\frac{3\log(2)-2}{4}\tag{10} \end{align} $$


Thus, we have $$ \begin{align} &3\int_0^1\left[\frac14\left(\frac1{2(1+t)}-\frac13\right)+\frac1{2(1+t)^2}\left(\frac{1+t}4-\frac13\right)\right]\,\mathrm{d}t\\[6pt] &=\frac{3\log(2)-2}{4}\\[6pt] &=\sum_{k=1}^\infty\frac1{(4k-1)4k(4k+1)} \end{align} $$ Perhaps there is another way to compute the area of the deltoid that makes it turn into the sum. However, it looks like a coincidence from this approach.

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  • $\begingroup$ A pretty moving diagram, and I had not spotted the midpoint point before so +1, but sadly no direct connection between the two ways of reaching $\frac{3}{4} \log_e(2) - \frac{1}{2}$ $\endgroup$ – Henry Apr 11 '14 at 22:59

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