# Chaitanya's Random Pages

## March 28, 2015

### A triangle centre arising from central projections

Filed under: mathematics — ckrao @ 10:23 pm

In this recent post, the following figure was formed by considering two triangular regions in three-dimensional space, where $a,b,c > 0$: $\displaystyle ax + by + cz = 1, 0 \leq x, 0 \leq y, 0 \leq z,$ $\displaystyle ax + by + cz = 1, x \leq 1, y \leq 1, z \leq 1.$ The intersection of the two triangles is the cross-section of a cube, but in this post we wish to explore further the centre of similarity of the two triangles.

The line joining $(1/a, 0, 0)$ and $((1-b-c)/a, 1, 1)$ satisfies \begin{aligned} (x,y,z) &= (1/a, 0, 0) + t\left[ ((1-b-c)/a, 1, 1)-(1/a, 0, 0) \right]\\ &= (t,t,(1-ta-tb)/c), t \in \mathbb{R}. \end{aligned}

Similarly, the line joining $(0,1/b,0)$ and $(1,(1-a-c)/b,1)$ satisfies $(x,y,z) = (u, (1-ua-uc)/b, u), u \in \mathbb{R}.$

Equating the two expressions gives $t = u$ and $t = (1-ta-tc)/b$ from which $t = u = 1/(a+b+c)$. The point of intersection is therefore at $(t,t,(1-ta-tb)/c) = (1/(a+b+c), 1/(a+b+c), 1/(a+b+c))$. By symmetry of this expression the line joining $(0,0,1/c)$ and $(1,1,(1-a-b)/c)$ also passes through this point. This is the point on the plane $ax + by + cz = 1$ that is equi-distant from the xy-, yz- and xz- coordinate planes. It is also the central projection of the origin onto the plane along the vector parallel to $(1,1,1)$.

In terms of the original two triangles this point is neither the centroid, incentre, orthocentre, circumcentre nor other commonly encountered triangle centre. Let us find the barycentric coordinates of this point (call it $S$) in terms of the triangle with vertices at $A(1/a,0,0), B(0,1/b,0), C(0,0,1/c)$. The first barycentric coordinate will be the ratio of the area of $\triangle SBC$ to the area of $\triangle ABC$. Since $B$ and $C$ have the same x-coordinate, this will be the ratio of the x-coordinates of $S$ to $A$, which is $1/(a+b+c) / (1/a) = a/(a+b+c)$. By symmetry it follows that the barycentric coordinates have the attractive form $\displaystyle \frac{a}{a+b+c} : \frac{b}{a+b+c} : \frac{c}{a+b+c}.$

Let the side lengths of $\triangle ABC$ be $BC = x, CA = y, AB = z$. Then by Pythagoras’ theorem, $x^2 = 1/b^2 + 1/c^2, y^2 = 1/a^2 + 1/c^2, z^2 = 1/a^2 + 1/b^2$. Hence $x^2 + y^2 - z^2 = 2/c^2$.

By the cosine rule, $x^2 + y^2 - z^2 = 2xy \cos C$ (where $C = \angle ACB$) which equals $2/c^2$ from the above expression. Therefore $c^2 = \sec C/xy$ and similarly we obtain $a^2 = \sec A/yz, b^2 = \sec B/xz$. Then \begin{aligned} \frac{a}{a+b+c} &= \frac{\sqrt{\sec A}/\sqrt{yz}}{\sqrt{\sec B}/\sqrt{xz} + \sqrt{\sec C}/\sqrt{xy} + \sqrt{\sec A}/\sqrt{yz}}\\ &= \frac{\sqrt{x\sec A}}{\sqrt{x\sec A} + \sqrt{y\sec B} + \sqrt{z\sec C}}.\end{aligned}

By the sine rule, $x = 2R\sin A$ ( $R$ being the circumradius of $\triangle ABC$) from which $x \sec A = 2R\sin A/\cos A = 2R\tan A$. Hence the barycentric coordinates of $S$ may be written in non-normalised form as $\displaystyle {\sqrt{\tan A}} : {\sqrt{\tan B}} : {\sqrt{\tan C}}.$

Comparing this with the coordinates of the orthocentre $\tan A : \tan B : \tan C$, the point $S$ is known as the square root of the orthocentre (see Theorem 1 of ). Note that the real existence of the point requires $\triangle ABC$ to be acute, which it is when $a,b,c > 0$. A geometric construction of the square root of a point is given in Section 8.1.2 of .

#### References

 Miklós Hoffmann, Paul Yiu. Moving Central Axonometric Reference Systems, Journal for Geometry and Graphics, Volume 9 (2005), No. 2, 127–134.

 Paul Yiu. “Introduction to the Geometry of the Triangle”. http://math.fau.edu/Yiu/GeometryNotes020402.pdf
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