Problem of the week #84 - November 4th, 2013

[Jameson]

Given a point $A$ that outside a circle so that $AT$ is tangent to the circle in point $T$
And $AC$ is a secant to that circle in points $B,C$.

From points $B,C$ we build heights to $AT$ in points $D,E$. (point $E$ is between points $D,T$)

$BD=b$, $CE=c$, angle $TAC=\alpha$.

What's the expression of radius $r$ in terms of $b,c,\alpha$?
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[Jameson]

Congratulations to the following members for their correct solutions:

1) MarkFL
2) anemone
3) Opalg

Solution (from MarkFL):

Spoiler

Please refer to the following diagram:

Let the center of the circle be oriented at the origin of an $xy$-coordinate system. Let the line $\overline{AC}$ lie along the line given by:

\(\displaystyle y=\cot(\alpha)x+k\)

Substituting for $y$ into the equation of the circle, we obtain:

\(\displaystyle x^2+\left(\cot(\alpha)x+k \right)^2=r^2\)

Expanding, we find:

\(\displaystyle \left(1+\cot^2(\alpha) \right)x^2+2k\cot(\alpha)x+k^2-r^2=0\)

Using the Pythagorean identity \(\displaystyle \csc^2(\theta)=\cot^2(\theta)+1\) we may write:

\(\displaystyle \csc^2(\alpha)x^2+2k\cot(\alpha)x+k^2-r^2=0\)

Multiplying through by \(\displaystyle \sin^2(\alpha)\) we get:

\(\displaystyle x^2+2k\sin(\alpha)\cos(\alpha)x+ \sin^2(\alpha)\left(k^2-r^2 \right)=0\)

Applying the quadratic formula, the roots of this quadratic are then given by:

\(\displaystyle x=\frac{-2k\sin(\alpha)\cos(\alpha)\pm \sqrt{\left(2k\sin(\alpha)\cos(\alpha) \right)^2-4(1) \left(\sin^2(\alpha)\left(k^2-r^2 \right) \right)}}{2\cdot1}\)

Simplifying, we obtain:

\(\displaystyle x=\frac{-2k\sin(\alpha)\cos(\alpha) \pm2\sin(\alpha)\sqrt{\left(k\cos(\alpha) \right)^2- \left(k^2-r^2 \right)}}{2}\)

\(\displaystyle x=\sin(\alpha)\left(-k\cos(\alpha)\pm\sqrt{k^2\cos^2(\alpha)-k^2+r^2} \right)\)

\(\displaystyle x=\sin(\alpha)\left(-k\cos(\alpha)\pm\sqrt{r^2-k^2\sin^2(\alpha)} \right)\)

The smaller of these roots is the $x$-coordinate of point $C$ and the larger is the $x$-coordinate of point $B$, hence:

\(\displaystyle b=r-\sin(\alpha)\left(-k\cos(\alpha)+\sqrt{r^2-k^2\sin^2(\alpha)} \right)=r+\sin(\alpha)\left(k\cos(\alpha)-\sqrt{r^2-k^2\sin^2(\alpha)} \right)\)

\(\displaystyle c=r-\sin(\alpha)\left(-k\cos(\alpha)-\sqrt{r^2-k^2\sin^2(\alpha)} \right)=r+\sin(\alpha)\left(k\cos(\alpha)+\sqrt{r^2-k^2\sin^2(\alpha)} \right)\)

Adding these two equations, we get:

\(\displaystyle b+c=2r+2k\sin(\alpha)\cos(\alpha)\)

Solving this for $k$, there results:

\(\displaystyle k=\frac{b+c-2r}{2\sin(\alpha)\cos(\alpha)}\)

Substituting for $k$ into the expression equal to $b$, we find:

\(\displaystyle b=r+\sin(\alpha)\left(\left(\frac{b+c-2r}{2\sin(\alpha)\cos(\alpha)} \right)\cos(\alpha)-\sqrt{r^2-\left(\frac{b+c-2r}{2\sin(\alpha)\cos(\alpha)} \right)^2\sin^2(\alpha)} \right)\)

Simplifying, we find:

\(\displaystyle b=\frac{b+c}{2}-\sin(\alpha)\sqrt{r^2-\left(\frac{b+c-2r}{2\cos(\alpha)} \right)^2}\)

\(\displaystyle \sqrt{r^2-\left(\frac{b+c-2r}{2\cos(\alpha)} \right)^2}=\frac{c-b}{2\sin(\alpha)}\)

Squaring:

\(\displaystyle r^2-\left(\frac{b+c-2r}{2\cos(\alpha)} \right)^2=\left(\frac{c-b}{2\sin(\alpha)} \right)^2\)

Multiply through by \(\displaystyle 4\cos^2(\alpha)\):

\(\displaystyle 4\cos^2(\alpha)r^2-\left(b+c-2r \right)^2=\left((c-b)\cot(\alpha) \right)^2\)

\(\displaystyle 4\cos^2(\alpha)r^2-(b+c)^2+4r(b+c)-4r^2=\left((c-b)\cot(\alpha) \right)^2\)

\(\displaystyle 4\sin^2(\alpha)r^2-4(b+c)r+(b+c)^2+\left((c-b)\cot(\alpha) \right)^2=0\)

\(\displaystyle 4\sin^2(\alpha)r^2-4(b+c)r+b^2+2bc+c^2+\cot^2(\alpha)\left(c^2-2bc+b^2 \right)=0\)

\(\displaystyle 4\sin^2(\alpha)r^2-4(b+c)r+\csc^2(\alpha)\left(b^2+c^2 \right)+2bc\left(1-\cot^2(\alpha) \right)=0\)

Applying the quadratic formula, and taking the smaller of the two roots, we obtain:

\(\displaystyle r=\frac{4(b+c)-\sqrt{16(b+c)^2-16\sin^2(\alpha)\left(\csc^2(\alpha)\left(b^2+c^2 \right)+2bc\left(1-\cot^2(\alpha) \right) \right)}}{2\left(4\sin^2(\alpha) \right)}\)

\(\displaystyle r=\frac{b+c-\sqrt{(b+c)^2-\left(\left(b^2+c^2 \right)-2bc\cos(2\alpha) \right)}}{2\sin^2(\alpha)}\)

\(\displaystyle r=\frac{b+c-\sqrt{2bc\left(1+\cos(2\alpha) \right)}}{2\sin^2(\alpha)}\)

\(\displaystyle r=\frac{b+c-2\cos(\alpha)\sqrt{bc}}{2\sin^2(\alpha)}\)