# Eritrea's Theorem

According to this newspaper, an Eritrean high school student named Saied Mohammed Ali has discovered a new geometric theorem. Another source seems to say that it's the following:

Say you have a triangle, with sides of length $a$, $b$, and $c$. Draw the medians (lines $\overline{AG}$, $\overline{BI}$, and $\overline{CH}$ in the diagram), and the altitudes (lines $\overline{AD}$, $\overline{BF}$, and $\overline{CE}$ in the digram).

Call the distance between where the median and altitude hit a given side the sma of that side. In the diagram, the smas are $\overline{GD}$, $\overline{IF}$, and $\overline{HE}$.

Call the length of the sma on side $a$, $\alpha$. Similarly, on sides $b$ and $c$ we have smas $\beta$ and $\gamma$.

The theorem is: $$a\alpha+b\beta=c\gamma$$

In the picture, we have $5.19\times0.09+4.28\times0.9=4.39\times0.98$, which is true up to rounding error.

How would you prove this? I have almost no experience in geometry, so I wouldn't even know where to start on this. Thanks! • This theorem seems to be a major source of Eritrean pride. – Akiva Weinberger Apr 1 '15 at 1:42
• How does one distinguish between side C and the other sides? – Mark Viola Apr 1 '15 at 1:44
• @Dr.MV With lack of a better term (that I know of), let's call $D,F,$ and $E$ altitude bases, and $G,I,H$ median bases. With sides $A$ and $B$, in the diagram, if you go clockwise, you encounter a median base and then an altitude base. With side $C$, the order is reversed — it's an altitude base and then a median base. (Cont'd) – Akiva Weinberger Apr 1 '15 at 1:46
• @Dr.MV I guess if the asymmetry really bothers you, we can work with signed distances, and get the equality $a\alpha+b\beta+c\gamma=0$, where a sma is positive if you get a median base and then an altitude base going clockwise, and negative if you get them in the reverse order. (If they coincide, the sma has length $0$, clearly.) – Akiva Weinberger Apr 1 '15 at 1:47
• @Dr. MV: According to tesfanews.net/eritreas-theorem "the product of the middle side and its sma is equal to the sum of the products of the remaining sides and their respective smas (the sum of the products of the length of the longest side and its sma and the length of the shortest side and its sma)". – g.kov Apr 1 '15 at 4:55

This is a different way to obtain the key equality $2a\alpha = c^2-b^2$.

We will use notion of power of a point with respect to a circle.

On the one hand, power of $G$ with respect to a circle centered at $I$ and radius $\frac b2$ equals $GI^2-\left(\frac b2\right)^2=\frac{c^2-b^2}{4}$.

On the other hand, this circle passes through $C$ and $D$ so the power of $G$ is $GD \cdot GC = \alpha \cdot \frac a2$ (assuming segments are oriented).

Thus $\frac{c^2-b^2}{4}=\alpha \cdot \frac a2$, so $2a \alpha = c^2-b^2$.

• Picture – Akiva Weinberger Apr 1 '15 at 11:21
• – Akiva Weinberger Apr 1 '15 at 11:34
• I'm accepting yours because it seems the simpler of the two. – Akiva Weinberger Apr 1 '15 at 11:37
• @columbus8myhw: I haven't read the article in full, but one paragraph talks about applications in a broad variety of fields. So perhaps the price was awarded not because this thing is difficult to prove, but instead because it is useful in some way, and was sufficiently hard to find that it hadn't been knon so far. I don't know. I'd not at all be surprised if these “new theorems” were to turn up in some forgotten 19th century work of some well known mathematician. I've seen this happen before. – MvG Apr 1 '15 at 11:45
• Very little of mathematics has 'applications in a broad variety of fields' and this theorem probably won't be 'useful' in any way at all. This is a nice result because it's sometimes assumed that all the true, easily understandable statements about triangles have already been made. The ease or otherwise of its proof doesn't matter. – user117644 Apr 1 '15 at 13:01

# Express α, β, γ in terms of a, b, c

According to the Encyclopedia of Triangle Centers, the orthocenter $X(4)$ has barycentric coordinates $[\tan A:\tan B:\tan C]$. From the cosine law you have $\cos C=\frac{a^2+b^2-c^2}{2ab}$ and likewise for the other angles. So you get

\begin{align*} \tan C&=\frac{\sin C}{\cos C}=\frac{\sqrt{1-\cos C^2}}{\cos C} =\frac{\sqrt{4a^2b^2-(a^2+b^2-c^2)^2}}{a^2+b^2-c^2} \\&=\frac{\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}}{a^2+b^2-c^2} \\&=\frac{4V}{a^2+b^2-c^2} \end{align*}

where $V$ denotes the area of the triangle, as obtained from Heron's formula. By canceling the $4V$ term and multiplying with all the denominators (which is allowed for homogeneous coordinates), you might write the barycentric coordinates of $X(4)$ also as

$$\begin{bmatrix} (a^2-b^2+c^2)(a^2+b^2-c^2)\\ (b^2-c^2+a^2)(b^2+c^2-a^2)\\ (c^2-a^2+b^2)(c^2+a^2-b^2) \end{bmatrix}$$

You get the footpoints of the heights by setting one of these coordinates to zero. So for example $D$ has barycentric coordinates

$$\begin{bmatrix} 0\\ (b^2-c^2+a^2)(b^2+c^2-a^2)\\ (c^2-a^2+b^2)(c^2+a^2-b^2) \end{bmatrix}\sim\begin{bmatrix} 0\\ b^2-c^2+a^2\\ c^2+a^2-b^2 \end{bmatrix}$$

which means you can write its Euclidean coordinates as

\begin{align*} D&=\frac{b^2-c^2+a^2}{(b^2-c^2+a^2)+(c^2+a^2-b^2)}B +\frac{c^2+a^2-b^2}{(b^2-c^2+a^2)+(c^2+a^2-b^2)}C \\&=\frac{a^2+b^2-c^2}{2a^2}B + \frac{a^2-b^2+c^2}{2a^2}C \\&=B + \frac{a^2-b^2+c^2}{2a^2}(C-B) \\&=B + \left(\frac12 + \frac{-b^2+c^2}{2a^2}\right)(C-B) \\&=G + \frac{-b^2+c^2}{2a^2}(C-B) \end{align*}

The distance between $D$ and $G$ is

$$\alpha=\lVert D-G\rVert = \left\lvert\frac{-b^2+c^2}{2a^2}\right\rvert\cdot\lVert C-B\rVert = \frac{\lvert b^2-c^2\rvert}{2a}\\ 2a\alpha=\lvert b^2-c^2\rvert$$

# Orientation and order

Up to here, the above is an alternative to the shorter deduction timon92 posted in his answer. The discussion which follows below applies no matter how one obtains that formula for $2a\alpha$.

Using this formula, the equation of the theorem (multiplied by $2$ to simplify things) would be

$$\lvert b^2-c^2\rvert + \lvert c^2-a^2\rvert = \lvert a^2-b^2\rvert$$

This is not always the case. But if you use signed distances, e.g. always measured in counter-clockwise direction, you can omit the absolute values. Then write the formula as “sum equals zero” as columbus8myhw suggests in his comment, and you obtain

$$(b^2-c^2) + (c^2-a^2) + (a^2-b^2) = 0$$

which is obviously true.

If you prefer unsigned distances, when does the equation with those hold? It holds if and only if the difference inside the absolute value function has equal sign for both terms on the left hand side of the equation but opposite sign on the right. So you have two cases to consider:

That's what the “middle side” g.kov quoted in his comment refers to: the $c$ on the right hand side of the equation must be the side of median length.

• The equality $2a\alpha=|b^2-c^2|$ can be obtained much more easily. It is enough to observe that $\alpha \cdot \frac a2$ is the power of point $G$ with respect to a circle centered at $I$ passing through $A,C,D,E$. – timon92 Apr 1 '15 at 9:51
• @timon92: For me, homogeneous coordinates, triangle centers and polynomials in the edge lengths are closer to my daily work than powers of a point. But I would like to see that deduction. Do you want to post an answer on that? I think it would nicely complement mine. – MvG Apr 1 '15 at 10:01
• I posted an answer. – timon92 Apr 1 '15 at 10:17

After having reviewed the paper thoroughly, I revealed some important concept from the basic equation of the theorem. I am still working on the paper. $$\frac1{\tan\beta} = \frac1{\tan\alpha}+\frac1{\tan\theta},$$ where the angles $\beta, \alpha$ and $\theta$ are the angles that are formed respectively when the medians drawn to the middle side, the smallest side and the longest side are constructed. Therefore, this above stated equation is very helpful to draw the three medians. The theorem is very easy, basic and important as well.

• I derived the above relation directly from the Eritrea's theorem, posted here,months ago. – math Aug 17 '15 at 18:33
• How does the above stated result hold true for an equilateral triangle? How would you prove it? – math Sep 4 '15 at 19:41
• How does the above stated result hold true for an equilateral triangle? How would you prove this? – math Sep 4 '15 at 19:46

My friend let me that, he see the theorem appeared early from a book in Viet Nam, Author of the book is Nguyen Minh Ha, Publish since 2000. See vidu 5.22 I propose a generalization of the Eritrea theorem: 