A Touching Problem

Geometry Level 5

Quadrilateral $EFGH$ is a rectangle such that sides $EF$ and $GH$ are parallel to $x$ -axis and sides $FG$ and $HE$ are parallel to $y$ -axis.

An ellipse with equation $8x^2+20xy+23y^2-220x-380y+1725 = 0$ is inscribed in the rectangle such that it touches the sides at points $A$ , $B$ , $C$ , and $D$ .

Segment $AC$ and $BD$ intersect at point $O$ .

If $\angle AOB = \theta$ and $\tan(\theta) = \dfrac{a}{b}$ , where $a$ and $b$ are coprime positive integers, find $b-a$ .

The answer is 113.

1 solution

A Former Brilliant Member
May 27, 2017

For a general 2 degree curve $E=ax^2+2hxy+by^2+2gx+2fy+c = 0$ .

$\displaystyle \frac{\partial E}{\partial x} = 0$ represents the diameter of the curve such that the tangents at the endpoints of this diameter are parallel to $x \ axis$ .

Similarly, $\displaystyle \frac{\partial E}{\partial y} = 0$ represents the diameter of the curve such that the tangents at the endpoints of this diameter are parallel to $y \ axis$ .

Now let, $S = 8x^2+20xy+23y^2-220x-380y+1725 = 0$ .

Thus equation of line $AC$ is given by $\displaystyle \frac{\partial S}{\partial x} = 16x+20y-220 = 0$ .

Similarly equation of line $BD$ is given by $\displaystyle \frac{\partial S}{\partial y} = 20x+46y-380 = 0$ .

Thus, slopes of line $AC$ and line $BD$ are $\displaystyle \frac{-4}{5}$ and $\displaystyle \frac{-10}{23}$ .

Angle between lines $(\alpha)$ having slopes $m_1$ and $m_2$ is given by $\displaystyle\tan(\alpha) = \Bigg|\frac{m_1-m_2}{1+m_1m_2}\Bigg|$ .

Therefore, angle between line $AC$ and line $BD$ is given by

$\displaystyle\tan(\theta) = \Bigg|\frac{\displaystyle\big(\frac{-10}{23}\big)-\big(\frac{-4}{5}\big)}{\displaystyle1+\big(\frac{-10}{23}\big)\big(\frac{-4}{5}\big)}\Bigg| = \Bigg|\frac{42}{155}\Bigg|$ .

Thus $b = 155$ and $a = 42$ .

Therefore, $b-a = 155-42 = 113$ .

Bonus Question: Try proving the fact that I stated at the beginning of the solution.

Proof of the first two statements in the solution given by Sid Naik -------

For general 2 degree curves $E=ax^2+2hxy+by^2+2gx+2fy+c=0$ diameter is defined as Locus of midpoints of parallel chords .

Let the slope of parallel chords be $m$ .

Then consider the point $(x_1,y_1)$ .

Equation of chords whose midpoint is $(x_1,y_1)$ is given by $T=S_1$ .

Thus equation of chord is $axx_1+h(xy_1+yx_1)+byy_1+g(x+x_1)+f(y+y_1)+c=ax_1^2+2hx_1y_1+by_1^2+2gx_1+2fy_1+c$ .

Therefore, $x(ax_1+hy_1+g)+y(hx_1+by_1+f)-ax_1^2-2hx_1y_1-by_1^2-gx_1-fy_1=0$ .

Thus slope of chord is $\displaystyle-\frac{ax_1+hy_1+g}{hx_1+by_1+f}=m$ .

Thus $(ax_1+hy_1+g)+m(hx_1+by_1+f)=0$ .

But $(ax_1+hy_1+g)+m(hx_1+by_1+f)=0$ is locus of $(x_1,y_1)$ which is diameter of the curve.

Also $(ax+hy+g)=\displaystyle\frac{\partial E}{\partial x}$ and $(hx+by+f)=\displaystyle\frac{\partial E}{\partial y}$

Thus equation of diameter is given by $\displaystyle\frac{\partial E}{\partial x}+m\frac{\partial E}{\partial y}=0$ where slope of chords is $m$ .

Thus when chord (or tangent) is parallel to $x \ axis$ (i.e. $m=0$ ) equation of dameter is given by $\displaystyle\frac{\partial E}{\partial x}=0$ .

And when chord (or tangent) is parallel to $y \ axis$ (i.e. $m=\infty$ ) equation of diameter is given by $\displaystyle\frac{\partial E}{\partial y}=0$ .

Rohit Sharma - 4 years ago

A Touching Problem

The answer is 113.

1 solution

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