# Constructing a triangle given three concurrent cevians?

Well, I've been taught how to construct triangles given the $3$ sides, the $3$ angles and etc. This question came up and the first thing I wondered was if the three altitudes (medians, concurrent$^\text {any}$ cevians in general) of a triangle are unique for a particular triangle.

I took a wild guess and I assumed Yes! Now assuming my guess is correct, I have the following questions

How can I construct a triangle given all three altitudes, medians or/and any three concurrent cevians, if it is possible?

N.B: If one looks closely at the order in which I wrote the question(the cevians coming last), the altitudes and the medians are special cases of concurrent cevians with properties

• The altitudes form an angle of $90°$ with the sides each of them touch.
• The medians bisect the side each of them touch.

With these properties it will be easier to construct the equivalent triangles (which I still don't know how) but with just any concurrent cevians, what other unique property can be added to make the construction possible? For example the angle they make with the sides they touch ($90°$ in the case of altitudes) or the ratio in which they divide the sides they touch ($1:1$ in the case of medians) or any other property for that matter.

EDIT

What André has shown below is a perfect example of three concurrent cevians forming two different triangles, thus given the lengths of three concurrent cevians, these cevians don't necessarily define a unique triangle. But also note that the altitudes of the equilateral triangle he defined are perpendicular to opposite sides while for the isosceles triangle, the altitude is, obviously also perpendicular to the opposite side with the remaining two cevians form approximately an angle of $50°$ with each opposite sides. Also note that the altitudes of the equilateral triangle bisects the opposite sides and the altitude of the isosceles triangle bisects its opposite sides while the remaining two cevians divides the opposite sides, each in the ratio $1:8$.

Now, given these additional properties (like the ratio of "bisection" or the angle formed with opposite sides) of these cevians, do they form a unique triangle (I'm assuming yes on this one) and if yes, how can one construct that unique triangle with a pair of compasses, a ruler and a protractor?

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You noticed that medians, altitudes, bisectors, etc are all cevians. So how can a set of cevians be unique to a triangle? You generally have freedom of choice in where the cevians intersect. –  EuYu Oct 30 '12 at 15:01
@EuYu why can't the set of all concurrent cevians of a triangle be unique? –  user31280 Oct 30 '12 at 15:05
Medians are a set of concurrent cevians. So are altitudes. There's no guarantee that a set of medians for one triangle cannot be a set of altitudes or bisectors for another. This holds even more true for arbitrary cevians. –  EuYu Oct 30 '12 at 15:08
@ I agree with you, although I would have preferred an example to back it up. But given three altitudes, we know they are concurrent and they form an angle of $90°$ with the sides they touch. Are these altitudes unique? –  user31280 Oct 30 '12 at 15:13
The sides of a triangle can be calculated from the medians (e.g. using Apollonius' theorem). Similarly the sides of a triangle can be calculated from the altitudes (e.g. via the area). –  Henry Oct 30 '12 at 15:33

It is clear that the lengths of concurrent cevians cannot always determine the triangle. Indeed, they probably never can. But if it is clear, we must be able to give an explicit example.

Cevians $1$: Draw an equilateral triangle with height $1$. Pick as your cevians the altitudes.

Cevians $2$: Draw an isosceles triangle $ABC$ such that $AB=AC$, and $BC=\dfrac{10}{12}$, and the height of the triangle with respect to $A$ is equal to $1$. Then $AB=AC=\sqrt{1+(5/12)^2}=\dfrac{13}{12}$.

There are (unique) points $X$ and $Y$ on $AB$ and $AC$ respectively such that $BY=CX=1$. This is because as a point $T$ travels from $B$ to $A$ along $BA$, the length of $CT$ increases steadily from $\dfrac{10}{12}$ to $\dfrac{13}{12}$, so must be equal to $1$ somewhere between $B$ and $A$. Let $X$ be this value of $T$.

Let one of our cevians be the altitude from $A$, and let $BY$ and $CX$ be the other two cevians. These three cevians are concurrent because of the symmetry about the altitude from $A$, and they all have length $1$.

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Thanks! please read the last part of my question. –  user31280 Oct 31 '12 at 14:36
Some further geometric constraints do, as we know, yield uniqueness. Categorizing the ones that do could be painful. I agree that division ratio should be one of them. For Euclidean constructibility, I would cheat and go immediately to coordinates. –  André Nicolas Oct 31 '12 at 14:58

I seem to recall reading that two of the familiar sets of cevians --@Henry's comment to OP suggests that I must be thinking of altitudes and medians-- are easily proven to determine a triangle; but the last --angle bisectors-- is tricky (but still determinative?). Or maybe I'm thinking of something else entirely. Edit: @Beni's answer suggests that I'm remembering properly.

Well, here's a coordinate proof for the case of altitudes:

Consider altitudes of length $a$, $b$, $c$ dropped from respective vertices $A$, $B$, $C$. Place $C$ at $(0,c)$, vertex $A$ at $(-p,0)$, and vertex $B$ at $(q,0)$ for non-negative $p$ and $q$.

The equations of lines $AC$ and $BC$ are $$AC: \quad -\frac{x}{p}+\frac{y}{c}-1=0 \qquad\qquad BC: \frac{x}{q}+\frac{y}{c}-1=0$$

Since $a$ is the distance from $A$ to $BC$, and $b$ the distance from $B$ to $AC$,

$$a = \frac{|-\frac{p}{q}-1|}{\sqrt{\frac{1}{q^2}+\frac{1}{c^2}}}=\frac{c \left(p+q\right)}{\sqrt{q^2+c^2}} \qquad b = \frac{c(p+q)}{\sqrt{p^2+c^2}}$$

We can solve this system for $p$ and $q$, getting

$$p = \frac{c\left(a^2b^2-b^2c^2+c^2a^2\right)}{d} \qquad q = \frac{c\left(a^2b^2+b^2c^2-c^2a^2\right)}{d}$$ where $$d^2 = \left( a b + b c + c a \right)\left(-a b + b c + c a \right) \left( a b - b c + c a \right)\left(a b + b c - c a\right)$$

Thus, a triple of altitudes determines a triangle (up to symmetry).

And here's one for medians:

Consider medians $a$, $b$, $c$ from respective vertices $A$, $B$, $C$. Place vertex $A$ at $(-p,0)$, vertex B at $(p,0)$, and vertex $C$ at $(c \cos t, c \sin t)$.

It's straightforward to compute the distance from $A$ to the midpoint of $BC$, and from $B$ to the midpoint of $AC$:

\begin{align} a^2 &= \left(-p-\frac{1}{2}(p+c\cos t)\right)^2+\left(0-\frac{1}{2}(c \sin t)\right)^2 = \frac{1}{4}\left(9p^2+c^2+6pc \cos t\right) \\ b^2 &= \frac{1}{4}\left(9p^2+c^2-6pc \cos t\right) \end{align}

Thus,

$$a^2 + b^2 = \frac{1}{2}\left( 9 p^2 + c^2 \right) \quad \implies \quad 9 p^2 = 2 a^2 + 2 b^2 - c^2$$

so that

$$4 a^2 = 2 a^2 + 2 b^2 + 6 p c \cos t \;\implies\; \cos^2 t = \frac{\left(a^2-b^2\right)^2}{c^2 \left( 2 a^2 + 2 b^2 - c^2 \right)}$$

giving us $p$ and $t$, and again determining a unique triangle (up to symmetry).

I'm not sure about the case of general cevians. I'm not even sure how one would articulate a general case.

Certainly, within a given triangle, a set of cevians corresponds to a triple of ratios $(\alpha,\beta,\gamma)$ with $\alpha\beta\gamma=1$. (That's Ceva's Theorem, after all.) So, perhaps you might ask,

For such a triple of ratios, does a triple of cevians $(a,b,c)$ uniquely determine a triangle?

Note, however, that the question doesn't include either the altitude case or the angle bisector case. For instance, we don't fix $(\alpha,\beta,\gamma)$ as we consider the universe of triangles with altitudes $(a,b,c)$.

Nevertheless, we can answer this version of the question in the affirmative. The proof is only slightly-modified from the median case:

We take the vertices $A(-p,0)$, $B(\gamma \, p,0)$, $C(c \cos t,c\sin t )$, and define cevian endpoints $D$ (on $BC$), $E$ (on $CA$), $F$ (on $AB$) such that

$$\alpha=\frac{|DC|}{|BD|} \qquad \beta=\frac{|EA|}{|AC|} \qquad \gamma=\frac{|FB|}{|AF|}$$

(Of course, $F$ is the origin.) A little coordinate algebra gives

\begin{align} a^2 &= |AD|^2 =\frac{1}{1+\alpha^2} \left(p^2(1+\alpha+\alpha\gamma)^2+2pc(1+\alpha+\alpha\gamma)\cos t+c^2\right)\\ b^2 &= |BE|^2 =\frac{1}{1+\beta^2} \left(p^2(1+\gamma+\beta\gamma)^2-2pc(1+\gamma+\beta\gamma)\cos t+c^2\right) \end{align}

Thus, we can eliminate $p\cos t$ from the system:

$$\frac{1+\gamma+\beta\gamma}{1+\beta^2} a^2 + \frac{1+\alpha+\alpha\gamma}{1+\alpha^2} b^2 = P p^2 + Q$$

for $P$ and $Q$ that I won't write down here. Clearly, there's a unique solution for non-negative $p$, which in turn gives a solution for $t$.

As for geometric construction ... The toolset "a pair of compasses, a ruler, and a protractor" seems haphazardly suggested: If you have a ruler and protractor --assuming infinite precision-- then you don't need a compass at all to draw a straight-line figure.

Classic straightedge-and-compass construction isn't always possible, as one could easily take $(a,b,c)$ to be a non-constructible trio of lengths --for instance, take $a$ to define the unit length, and have $b=\pi$ and $c=\sqrt[3]{2}$-- and/or similarly take the ratios $(\alpha, \beta, \gamma)$ to ensure non-constructible segments.

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For the angle bisectors, there is a theorem which says that for any given numbers $a,b,c>0$ there exists a unique triangle (up to isometry) with the lengths of angle bisectors equal to $a,b,c$, respectively.

The proof is quite long, and involves using a fixed point theorem. You can find it here, for example.

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You can find a nice solution to the case of altitudes here: http://www.flickr.com/photos/52771834@N00/5751939270/sizes/l/in/photostream/

You can also find nice solutions to the case of medians here: http://mathforum.org/library/drmath/view/55411.html or http://www.qc.edu.hk/math/Junior%20Secondary/Median.htm

As regards the case of cevians, let's solve the following problem:

See the figure A below:

Let $AB=c$, $BC=a$, $CA=b$, $AD=m$,$BE=n$,$CF=p$, $\gamma=\frac{AF}{FB}$,$\alpha=\frac{BD}{DC}$, and $\beta=\frac{CE}{EA}$, draw $\triangle ABC$ given $m$, $n$, $p$, $\alpha$ and $\beta$. Note that $\alpha$ and $\beta$ are defined as the ratio of measures of two line segments:

$$\alpha = \frac{a_1}{a_2} \quad (I)$$ and $$\beta = \frac{b_1}{b_2} \quad (II)$$

See the figure 1.

Using auxiliary lines and similarity we can deduce that $$\frac{DO}{OA}= \frac{1}{\gamma(1+\alpha)}, \quad (1)$$ $$\frac{FO}{OC}= \frac{\alpha \gamma}{(1+\gamma)}, \quad (2)$$ and $$\frac{EO}{OB}= \frac{\beta \gamma}{(1+\beta)}. \quad (3)$$

As $\beta=\frac{CE}{EA}$ it follows that: $$CE=\frac{\beta b}{1+ \beta}, \quad (4)$$ and $$EA=\frac{b}{1+ \beta}. \quad (5)$$ From Ceva's Theorem we get: $$\gamma \alpha \beta =1 \Rightarrow \gamma = \frac{1}{\alpha \beta}. \quad (6)$$ So with the values of $m$, $n$, $p$, $\alpha$ and $\beta$, we can calculate $\gamma$, $DO$, $OA$, $FO$, $OC$, $EO$, $OB$ using equations $(6)$, $(1)$, $(2)$ and $(3)$.

In addition if we use equation $(4)$, $(5)$ and Stewart's Theorem in $\triangle COA$ we get: $$b^2 = OA^2(1+ \beta) + OC^2 \frac{(1+\beta)}{\beta} - EO^2 \frac{(1+\beta)^2}{\beta}. \quad(7)$$ Now let's define four auxiliary variables $h_{OA}$,$h_{OC}$, $z$ and $y$: $$h_{OA}^2= OA \cdot (1+\beta)OA, \quad (8)$$ $$h_{OC}^2= OC \cdot (1+\beta^{-1})OC, \quad (9)$$ $$z^2= h_{OA}^2 + h_{OC}^2, \quad (10)$$ $$y^2= (1+ \beta^{-1})EO \cdot (1+\beta)EO. \quad (11)$$ Using $(7)$, $(8)$, $(9)$, $(10)$ and $(11)$ we get: $$b^2=z^2-y^2. \quad(12)$$

So to solve this problem we must:

1. Find out $\gamma$. (If we define $\gamma$ as $\gamma=\frac{c_1}{c_2}$, then Fig. 2 shows how to determine $c_1$ and $c_2$ based on Equations $(I)$, $(II)$ and $(6)$).

2. Find out the position of point $O$ in each cevian ($AD$, $CF$ and $BE$).The procedure is outlined in figure 3, based on equations $(1)$, $(2)$, $(3)$ and $(6)$.

3. Find out $h_{OA}$, $h_{OC}$, $y$, and $z$ as shown in figures 4 and 5.(Based on Equations $(8)$, $(9)$, $(10)$ and $(11)$).

4. Find out $b$ as shown in the last circle of figure 4.(As stated in Equation $(12)$).

5. Draw the $\triangle ABC$ as shown in the right part of figure 5, starting at $\triangle AOC$.

I know that it is a long resolution, but it is the only one I could find at least until now.

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