euclidean-geometrygeometryplane-geometry
In the figure, if $C$, $A$ and $N$ are points of tangency , determine $CM$. ($S:CM=2$)
I try
CNOI is a square
$\therefore CN = l\boxed{}\sqrt2 \implies l_\boxed{}=3\sqrt2\\
FN.MN = DN.IN \implies 9.(6+CM) = DN.3\sqrt2\\
\therefore 18+3CM = DN\sqrt2$
out-of-scale figure
$PG = r_1 - r, PR = r_1 + r$ and so by Pythagoras, it is easy to see that $RG = 2 \sqrt{r r_1}$
Now if $O$ is the center of the circle with radius $R$,
$AG = 2r_1 - r ~$ and $ |OG| = |AG - AO| = |2r_1 - r - R| ~ $
By Pythagoras, $OG^2 = OR^2 - RG^2$
$\implies (2r_1 - r - R)^2 = (R-r)^2 - 4 r r_1$
Solving, $R r = R r_1 - r_1^2 \implies R r = r_1 (R - r_1)$. But as $R = r_1 + r_2$,
$R r = r_1 r_2 \tag1$
Now $KQ = r_2 - x, QS = r_2 + x$ and we obtain $SK = 2 \sqrt{xr_2}$
$OK = OB - KB = R - (2r_2 - x)$
$OK^2 = (R - 2r_2 + x)^2 = OS^2 - SK^2 = (R-x)^2 - 4 xr_2$
Solving, $Rx = r_2 (R - r_2) = r_1 r_2 \tag2$
From $(1)$ and $(2)$, we conclude that $r = x$.
Here is a geometrical solution without much algebra.
For a right triangle, if you draw a line through the points of tangency of the incircle with the perpendicular sides, it does bisect the arcs of the circumcircle on both sides. In other words, $M$ and $N$ are midpoints of minor arcs $AB$ and $BC$ respectively. At the end of the answer, I have shown a proof.
With that, note that $\triangle BTM \sim \triangle NKB$. That leads to,
$\frac{r}{a} = \frac{b}{r} \implies r = \sqrt{ab}$
As $FM$ is perpendicular bisector of $AB$ and $FN$ is perpendicular bisector of $BC$,
$\frac{AB}{2} = \sqrt{ab} + \frac{a}{\sqrt2}$ and $\frac{BC}{2} = \sqrt{ab} + \frac{b}{\sqrt2}$
As we know $AB$ and $BC$ in terms of $a$ and $b$, we are done, for $S_{\triangle ABC} = \frac 12 \cdot AB \cdot BC$.
Proof of the property that I used in the above answer -
Say $M$ and $N$ are midpoints of the arcs $AB$ and $BC$ and segment $MN$ intersects $AB$ and $BC$ at $T$ and $K$ respectively.
$\angle BIN = 45^\circ + \angle A/2$
So, $\angle KPN = 90^\circ + \angle A/2$
Also, $\angle PNK = \angle C/2$
That leads to $\angle BKM = \angle PKN = 45^\circ$
Also note that $\angle INK = \angle ICK = \angle C / 2$ so $ICNK$ is cyclic and therefore $\angle KIN = \angle KCN = \angle A / 2$
That leads to $\angle IKM = \angle A / 2 + \angle C / 2 = 45^\circ$
$\angle BKI = \angle BKM + \angle IKM = 90^\circ$.
So $K$ must be point of tangency of incircle with side $BC$. Finally since $KI \parallel BT$ and $BT = BK = KI$, $T$ is the point of tangency of incircle with side $AB$.
Best Answer
$ CN = l\boxed{}\sqrt2 \implies l_\boxed{}=3\sqrt2\\ FN.MN = DN.IN \implies 9.(6+CM) = DN.3\sqrt2\\ \therefore 18+3CM = DN\sqrt2(I)\\ \triangle OFP: OF^2 = 9^2+(3\sqrt2)^2+2.9.3\sqrt2(cos135^o)\implies OF = 3\sqrt{17}\\ \triangle DNO \sim \triangle FPO: \frac{DN}{12}=\frac{3\sqrt2}{3} \implies DN = 12 \sqrt2\\ From(I):18+3CM =12\sqrt2.\sqrt2 \implies 3CM = 6 \therefore \boxed{CM = 2} $