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#### Jester

##### Well-known member
MHB Math Helper
My solution
This is going to be algebraic. I'm assuming someone will give a geometric answer. If we flip the picture, then we can come up with the equation of the second circle (the one tangent to the line)

$(x-3s)^2+(y-s)^2 = s^2$

and the equation of the straight line itself $y = 9 - \dfrac{9}{12}x$

Now if $(x_0,y_0)$ is the point on the circle whose tangent is the line then we can also come up with the tangent line using calculus, namely

$y - y_0 = - \dfrac{(x_0 - 3s)}{(y_0-s)}( x - x_0)$

As the two lines must be the same gives (eliminate $y$ and isolate the coefficients wrt $x$)

$\dfrac{3}{4}-{\dfrac {{ x_0}}{{ y_0}-s}}+3\,{\dfrac {s}{{ y_0}-s}}=0\;\;\;(1)$

$-9+{\it y_0}+{\dfrac {{{ x_0}}^{2}}{{ y_0}-s}}-3\,{\dfrac {s{ x_0}} {{\it y_0}-s}} = 0\;\;\;\;(2)$

Furthermore, we know that $(x_0,y_0)$ is on the circle so

$(x_0-3s)^2+(y_0-s)^2 = s^2\;\;\;(3)$

From (1) we find that $y_0 = -3s + \dfrac{4}{3} x_0$ and with this, from (2) we obtain

$x_0 = \dfrac{36}{25} s+\dfrac{108}{25}$ and with these two gives (3) as

${\frac {72}{25}}\, \left( s-2 \right) \left( 2\,s-9 \right) =0$

Clearly $s = 9/2$ is too large thus giving $s = 2$, the radius of the circle.

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#### Opalg

##### MHB Oldtimer
Staff member
I used trigonometry. If $\theta$ is the angle $BAC$, then $\sin\theta = 3/5$. If $t = \tan(\theta/2)$ then one of the half-angle formulas says that $\sin\theta = \dfrac{2t}{1+t^2} = \dfrac35$, from which $3t^2-10t+3=0.$ The solutions to that are $t=1/3$ and $t=3$. Clearly $t=3$ is too big, and so $\tan(\theta/2) = 1/3.$ But the line $AO_1$ bisects angle $BAC$, and therefore $O_1Q/AQ = 1/3$. So $AQ=3s$. Since $QC = O_1R = 3s$, it follows that $12 = AC = 6s$, and so $s=2$.

#### Albert

##### Well-known member
I will use geometry ,
maybe someone want to try it first

#### Opalg

##### MHB Oldtimer
Staff member
I will use geometry ,
maybe someone want to try it first
I believe that the dashed lines in the diagram give a clue. They divide the triangle ABC into three smaller triangles, whose areas are $\frac12(12s)$, $\frac12(9(3s))$ and $\frac12(15s)$ (using Pythagoras to get the hypotenuse of triangle ABC as 15). The area of the whole triangle is $\frac12(9\times12)$. Equating that to the sum of the three smaller areas easily gives the equation $27s=54$ from which $s=2$.

#### Albert

##### Well-known member
I believe that the dashed lines in the diagram give a clue. They divide the triangle ABC into three smaller triangles, whose areas are $\frac12(12s)$, $\frac12(9(3s))$ and $\frac12(15s)$ (using Pythagoras to get the hypotenuse of triangle ABC as 15). The area of the whole triangle is $\frac12(9\times12)$. Equating that to the sum of the three smaller areas easily gives the equation $27s=54$ from which $s=2$.
yes ,you got it 