Ellipse area

Let $\mathcal{E}$ be the Steiner inellipse of the triangle $NTU$ , where $N\;(6,6)$ , $T\;(-2,2)$ and $U\;(8,-2)$ are the pairwise intersections of the lines $x-2y=-6$ , $-4x-y=30$ and $2x+5y=6$ . Then $\mathcal{E}$ is tangential to the lines $NT$ , $TU$ , $UN$ at their midpoints $K\;(2,4)$ , $L\:(3,0)$ and $M\:(7,2)$ . The centre of $\mathcal{E}$ is the centroid $S\;(4,2)$ of the triangle $NTU$ .

The point diametrically opposite $M$ on $\mathcal{E}$ is $J\;(1,2)$ . The tangent to $\mathcal{E}$ at $J$ will be parallel to the tangent to $\mathcal{E}$ at $M$ , and so will have gradient $-4$ . Thus the equation of the gradient to $\mathcal{E}$ at $J$ is $4x + y = 6$ .

The point diametrically opposite $K$ on $\mathcal{E}$ is $A'\;(6,0)$ . The tangent to $\mathcal{E}$ at $A'$ will be parallel to the tangent to $\mathcal{E}$ at $K$ , and so will have gradient $\tfrac12$ . Thus the equation of the gradient to $\mathcal{E}$ at $A'$ is $-x + 2y = -6$ .

Thus $\mathcal{E}$ is the ellipse we want. The area of $\mathcal{E}$ is equal to $\tfrac{\pi}{3\sqrt{3}}$ times the area of $NTU$ . Since the triangle $NTU$ has area $36$ , the area of $\mathcal{E}$ is $\pi4\sqrt{3}$ , making the answer $4+3 = \boxed{7}$ .

The lines seem to form an incomplete hexagon, so maybe we can turn this ellipse into a circle by applying a linear transformation to make the hexagon regular.

The points $O,P,Q,R$ are $(2/3,10/3),(4/3,2/3),(14/3,-2/3),(22/3,2/3)$ respectively. Let's first shift $P$ to $(0,0)$ , so that these points become $(-2/3,8/3),(0,0),(10/3,-4/3),(6,0)$ . Now, let's draw a simpler regular hexagon which we'll want to transform into these points. The red hexagon fragment has side length 1, and so the coordinates of its points are ${-\frac12\choose\frac{\sqrt3}2},{0\choose0},{1\choose0},{\frac32\choose\frac{\sqrt3}2}$ We now want a linear transformation in the form of a $2\times2$ matrix that takes these vectors to ${-\frac23\choose\frac82},{0\choose0},{\frac{10}3\choose-\frac43},{6\choose0}$ With a small amount of working out, we find that the following matrix works: $M=\begin{pmatrix} \frac{10}3 & \frac2{\sqrt3} \\ -\frac43 & \frac4{\sqrt3} \end{pmatrix}$ Now, a circle inside this red hexagon has height $\sqrt3$ , so it has radius $\frac{\sqrt3}2$ , and hence area $\frac34\pi$ . When we stretch the coordinate system from the red hexagon to the blue one, areas increase by a factor of $\det M$ . Hence, the ellipse in the question has area $\frac34\pi\det M=4\sqrt3\pi$

Find five tangent points

Example - $D$ intersection two diagonals $OQ$ and $PR$ and tangent point $L$ - intersection two lines $PQ$ and $ND$ .

$x = 3, y = 0$

$x = 6, y = 0$

$x = 1, y = 2$

$x = 7, y = 2$

$x = 2, y = 4$

Find ellipse equation for this five points $4x^2+2xy+7y^2-36x-36y+72=0$ and find his area.

I don't think I found a good approach here - hopefully someone will post a better one.

The equation of an ellipse in general position is $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$

where $B^2-4AC<0$ . For definiteness, we can set $A=1$ .

A line with equation $y=mx+c$ intersects the ellipse twice, once, or not at all. We're interested in tangents, which intersect exactly once.

Eliminating $y$ between these equations, $x^2+Bx(mx+c)+C(mx+c)^2+Dx+E(mx+c)+F=0$

This is a quadratic in $x$ . For this to have exactly one solution, we want the discriminant to be zero; ie $(Bc+2Cmc+D+Em)^2 - 4(1+Bm+Cm^2)(Cc^2+Ec+F) = 0$

The five given tangent lines give five pairs of values $(m,c)$ . Substituting these in, we get five equations in $B,C,D,E,F$ .

Solving these (I used a combination of Wolfram|Alpha and Excel Solver for this), we find the equation of the ellipse: $x^2+\frac12 xy+\frac74 y^2-9x-9y+18=0$

Helpfully, there is a formula to calculate the axes from this form, see for example here . Plugging in, we find the semi-axes have lengths $\sqrt{\frac23} \sqrt{11 \pm \sqrt{13}}$

so the area is $4\pi \sqrt3$ and the answer is $\boxed7$ .

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Just a note (this may help with a neater solution): the centre of the ellipse is $(4,2)$ . Translating the ellipse so its centre is the origin gives an easier form: $4x^2+2xy+7y^2-36=0$

This also makes finding the axes easier.