When an ellipse touches the sides of a triangle

Question

An ellipse touches the sides of a triangle $abc$ from inside in the points $a',b',c'$.

How can I prove, that the lines $ aa',bb',cc'$ meet in one point?

The ellipse equation is : $ \frac{x^2}{A^2} + \frac{x^2}{B^2}= 1 $

I've seen this kind of questions in old exams, therefore I would like to know how to handle such a proof right.

Do I have to build equations for $aa',bb',cc'$ ? If yes, how do do that?

Answer 1

A much simpler answer is obtained based on the fact that any triangle with any inscribed ellipse, can be transformed to a triangle where the inscribed ellipse is transformed into a circle. This circle, of course, is the incircle. It can be shown easily that the cevians (the line segments) joining the vertices to the opposites sides touching points of the incircle are concurrent, based on the properties of the incircle, and Ceva's theorem. A very to-the-point proof is given here.

Answer 2

I will first numerically/graphically verify the statement to be proved. Given $\triangle ABC$, it can be transformed into an isosceles right triangle, with vertices $L(0,0), M(1,0), N(0, 1)$ using the affine transformation $\mathbf{y} = \mathbf{A x + b}$, for some matrix $\mathbf{A}$ and some vector $\mathbf{b}$.

Clearly, an inscribed ellipse of the original triangle is transformed into an inscribed ellipse of the transformed triangle, and the line segments connecting the vertices to the touching points of the inscribed ellipse in the original triangle will be transformed into line segments in the transformed triangle. So if this set of the three line segments is concurrent, then so will be the set of the corresponding three line segments in the transformed triangle and ellipse, and vice versa.

From this, it is sufficient to show that the line segments connecting the vertices of the transformed triangle to the touching points of the transformed ellipse are concurrent, to show that they are concurrent in the original triangle.

Now suppose we are given the relative location of the touching points of the ellipse on sides $AB$ and $AC$, then these are transformed into known points $T_1$ and $T_2$ on $LM$ and $LN$ respectively, because the relative location of points on line segments does not change after an affine transformation, as defined above. The location of the touching point $T_3$ on side $MN$ is not arbitrary and can be found analytically.

Now we have three touching points, and three vertices, so we just connect them in pairs and see if they intersect at the same point. This is what I have done, and the figure below illustrates this. Thus I have verified that the line segments connecting vertices to the touching points of an in-ellipse on the opposite sides do indeed intersect in a single point.

To turn the above verification into a proof, let the touching point of the inscribed ellipse with the horizontal side be $T_1 = (t_1, 0)$ and the touching point with the vertical side be $T_2 = (0, t_2)$, and let the equation of the inscribed ellipse be

$ (r - r_0)^T Q (r - r_0) = 1$

where $r = [x,y]^T $,

Now the gradient of the ellipse at $T_1$ is given by

$ g_1 = 2 Q (T_1 - r_0) $

And we know that this vector will be pointing along $(-e_2) = - (0, 1)$, therefore,

$ Q (T_1 - r_0) = - \alpha e_2 , \hspace{10pt} \alpha \gt 0 $

From this

$ T_1 - r_0 = - \alpha Q^{-1} e_2 $

Since $T_1$ is on the ellipse, it follows that

$\alpha = \dfrac{1}{\sqrt{ e_2^T Q^{-1} e_2} } $

So that

$ T_1- r_0 = - \dfrac{ Q^{-1} e_2 }{ \sqrt{ e_2^T Q^{-1} e_2 } } $

Premultiplying this equation with $e_1^T $ and $e_2^T$ respectively, results in

$ e_1^T (T_1 - r_0) = t_1 - r_{0x} = - \dfrac{ e_1^T Q^{-1} e_2}{ \sqrt{ e_2^T Q^{-1} e_2 } } $

and

$ e_2^T (T_1 - r_0 ) = 0 - r_{0y} = - \sqrt{e_2^T Q^{-1} e_2 } $

Similarly, the gradient at $T_2$ is pointing in the negative $e_1$ direction, and applying the same method as above, we get

$ - r_{0x} = - \sqrt{ e_1^T Q^{-1} e_1 } $

and

$ t_2 - r_{0y} = - \dfrac{ e_2^T Q^{-1} e_1 }{ e_1^T Q^{-1} e_1 } $

Now let matrix $P = Q^{-1}$ , then

$ e_1^T Q^{-1} e_1 = P_{11} , e_1^T Q^{-1} e_2 = e_2^T Q^{-1} e_1 = P_{12} = P_{21} , e_2^T Q^{-1} e_2 = P_{22} $

Therefore, the equations we have now are

$t_1 - r_{0x} = - \dfrac{ P_{12} }{\sqrt{ P_{22} } } $

$ - r_{0y} = - \sqrt{ P_{22}} $

$ - r_{0x} = - \sqrt{P_{11}} $

$ t_2 - r_{0y} = - \dfrac{ P_{12} } { \sqrt{ P_{11} }} $

Eliminating $r_{0x}$ using the first and third equations, and eliminating $r_{0y}$ using the second and fourth equations, gives us,

$ t_1 = - \dfrac{ P_{12} }{\sqrt{ P_{22} } } + \sqrt{P_{11}} $

$ t_2 = - \dfrac{ P_{12} } { \sqrt{ P_{11} } } + \sqrt{P_{22}} $

And these two equations yield

$ t_1 \sqrt{P_{22} } = - P_{12} + \sqrt{P_{11}} \sqrt{P_{22}} $

$ t_2 \sqrt{P_{11}} = - P_{12} + \sqrt{P_{11}} \sqrt{P_{22}} $

Subtracting,

$ t_1 \sqrt{P_{22}} = t_2 \sqrt{P_{11} } $

Let $X = \sqrt{P_{11}} $, then

$P_{11} = X^2 $

$ P_{22} = \left( \dfrac{t_2}{t_1} \right)^2 X^2 $

$ P_{12} = - t_2 X + \left( \dfrac{t_2}{t_1} \right) X^2 $

and we also have

$ r_{0x} = \sqrt{ P_{11} } = X $

$ r_{0y} = \sqrt{ P_{22} } = \left(\dfrac{t_2}{t_1} \right) X $

What is left is to determine the value of $X$. For that, consider the unknown point $T_3$ which lies on side $MN$, the outward normal is

$ u = \dfrac{1}{\sqrt{2}} [ 1, 1]^T $

Applying the same method given above, we deduce that

$ T_3 - r_0 = \dfrac{ P u }{\sqrt{ u^T P u } } \hspace{25pt} (*) $

Premultiplying by $u^T $

$ u^T (T_3 - r_0) = u^T (T_3 - M + M - r_0) = 0 + \dfrac{1}{\sqrt{2}} \left( 1 - r_{0x} - r_{0y} \right) = \sqrt{ u^T P u } $

Using the above equations, this equation becomes

(where $\alpha = \dfrac{t_2}{t_1} $ )

$ \dfrac{1}{\sqrt{2}}( 1 - X - \alpha X ) = \sqrt{ \frac{1}{2}\left( X^2 + \alpha^2 X^2 + 2 ( - t_1 X + \alpha X^2 ) \right) } \hspace{25pt}(**) $

Squaring both sides of this equation gives

$ ( 1 + (1 + \alpha)^2 X^2 - 2 (1 + \alpha) X ) = X^2 (1 + \alpha)^2 - 2 t_1 X $

from this it follows that

$ X = \dfrac{1}{ 2 ( 1 + \alpha - t_1 ) } = \dfrac{ t_1}{ 2 (t_1 + t_2 - t_1 t_2 ) }$

Let $ d= 2 (t_1 + t_2 - t_1 t_2 ) $, then $ X = \dfrac{t_1}{d} $

thus we have shown that the center of the inscribed ellipse is

$ r_0 = [ \dfrac{t_1}{d} , \dfrac{t_2}{d} ] $

And

$ Q = P^{-1} = \bigg(\dfrac{1}{d^2} \begin{bmatrix} t_1^2 && t_1 t_2 (1 - d) \\ t_1 t_2 (1 - d) && t_2^2 \end{bmatrix} \bigg) ^{-1} = \dfrac{d}{ t_1^2 t_2^2 (2 - d) } \begin{bmatrix} t_2^2 && t_1 t_2 (d-1) \\ t_1 t_2 (d -1) && t_1^2 \end{bmatrix}$

To find the third tangency point, we use equation $(*)$, and we have

$ \dfrac{ P u}{\sqrt{u^T P u} } = \dfrac{[ t_1^2 + t_1 t_2 (1 - d) , t_2^2 + t_1 t_2 (1 - d) ]^T}{d^2 (1 - (1 + \alpha) X)}$

With $\alpha = \dfrac{t_2}{t_1} , X = \dfrac{t_1}{d} $ this becomes

$ \dfrac{ P u}{\sqrt{u^T P u} } = \dfrac{[ t_1^2 + t_1 t_2 (1 - d) , t_2^2 + t_1 t_2 (1 - d) ]^T}{ d (d - (t_1 + t_2)) }$

Hence,

$T_3 = r_0 + \dfrac{ P u}{\sqrt{u^T P u} } = [\dfrac{t_1}{d} , \dfrac{t_2}{d} ]^T + \dfrac{[ t_1^2 + t_1 t_2 (1 - d) , t_2^2 + t_1 t_2 (1 - d) ]^T}{ d (d - (t_1 + t_2)) }$

And the right hand side simplifies to

$T_3 = \dfrac{1}{d (d- t_1 - t_2)} [ t_1 (d - t_1 - t_2) + t_1^2 + t_1 t_2 (1 - d) , t_2 (d - t_1 - t_2) + t_2^2 + t_1 t_2 (1- d) ]^T $

And this simplifies further to the final form

$ T_3 = \dfrac{1}{d - t_1 - t_2} ( t_1 (1 - t_2), t_2 (1 - t_1) ) $

Now the intersection of $NT_1$ and $MT_2$ is easily shown to be

$ G = \dfrac{1}{1 - t_1 t_2} (t_1 ( 1- t_2), t_2 (1 - t_1) ) $

Clearly, point $G$ lies on $LT_3$. And this concludes the proof.

Answer 3

Lateral comment only, not an answer.

$$(x_c,y_c)=\left(\frac{t_1- t_1t_2}{1-t_1t_2},\frac{t_2-t_1 t_2}{1-t_1t_2}\right)$$

Answer 4

EDIT: the proof below works only for Steiner inscribed ellipse, not for any inscribed ellipse. Which makes this proof useless, sorry for that.

Depending the date of the "old exams" where you have seen this problem, it may have been solved by using affine transformations. Two people have already commented about that, so here are more details.

The construction is invariant by affine transformations, because it involves only operations that have this invariant: tangents, intersections. A triangle is transformed into a triangle, an ellipse is transformed into an ellipse, and as the ellipse inscribed into a triangle is unique, if a triangle T is transformed into a triangle U the ellipse inscribed in T gets transformed into the ellipse inscribed in U.

Then, there is an affine transformation that transforms the triangle $abc$ into a given equilateral triangle. So the intersection of $aa'$ with $bb'$ gets transformed into the intersection of $aa'$ with $bb'$ in the equilateral triangle, etc.

Proving the property for an equilateral triangle is easy, because the inscribed ellipse is a circle and the 3 lines intersect at the circle center. So the fact that the 3 lines intersect on only 1 point gets transformed back into the same property for the first triangle, because the number of intersections is an affine invariant. Proof is finished. And you don't have to explicit the equations.

It may seem handwavy, but given some geometry background that was current some 80 (?) years ago, it is perfectly valid. And the nice thing is, it applies to many constructions, especially with triangles and parallelograms.