How to find the relation between Area and Radius?

Question

Let $S$ be the circumcircle of a right triangle $ABC$ with $\measuredangle A = 90^{\circ}$ . Circle $X$ is a tangent to the lines $AB$ and $AC$ and internally to $S$ . Circle $Y$ is tangent to $AB$ and $AC$ and externally to $S$ . Prove that (radius of $X$) ·(radius of $Y$) is equal to four times the area of $\Delta ABC$.

My attempts:

The Circle X is touching the lines AB and AC and the Circle S, so I'm trying to calculate the relation between the radius of X and AB and AC, because half of product of AB and AC is the area of ABC, but so a relation between the radius of X and AB and AC can help. But I'm unable to find, I'm still thinking, Please help..

Answer 1

Let $\Delta ABC$ be a given triangle with $\measuredangle ACB=90^{\circ}$.

(I changed it. Sorry.)

Let $X$ be a center of the circle $X$, $x$ be a radius of the circle and $O$ is center of $S$.

Thus, in the standard notation $CX=x\sqrt2$, $OX=\frac{c}{2}-x$, $CO=\frac{c}{2}$ and $\measuredangle XCO=\left|45^{\circ}-\alpha\right|$.

Thus, by law of cosines we obtain: $$OX^2=CX^2+CO^2-2CX\cdot CO\cos\measuredangle XCO$$ or $$\left(\frac{c}{2}-x\right)^2=2x^2+\frac{c^2}{4}-cx\sqrt2\cos\left(45^{\circ}-\alpha\right)$$ or $$x=c\left(\sqrt2\cos\left(45^{\circ}-\alpha\right)-1\right).$$ By the same way we obtain $$y=c\left(\sqrt2\cos\left(45^{\circ}-\alpha\right)+1\right),$$ where $y$ is a radius of the circle $Y$.

Id est, $$xy=c^2\left(2\cos^2\left(45^{\circ}-\alpha\right)-1\right)=c^2\cos\left(90^{\circ}-2\alpha\right)=$$ $$=c^2\sin2\alpha=2c^2\sin\alpha\cos\alpha=4\cdot\frac{ab}{2}=4S_{\Delta ABC}.$$

Answer 2

Like @Michael, I'm also moving the right angle to $\angle C$. Also, let $$|BC| = 2a \qquad |CA| = 2b \qquad |AB| = 2c \qquad\text{so that}\quad |\triangle ABC| = \frac12(2a)(2b) = 2ab$$ The circumcenter of $\triangle ABC$ is the midpoint, $K$, of $\overline{AB}$; the circumradius is $c$.

Let $P$ be the center, and $r$ the radius, of the "internal" circle in question; note that $\overline{PC}$ is the diagonal of a square with sides aligned with legs of the original triangle. Let $\bigcirc P$ be tangent to $\bigcirc K$ at $T$; necessarily, $K$, $P$, $T$ are collinear. Finally, let $Q$ be such that $\triangle KPQ$ is a right triangle with hypotenuse $\overline{KP}$ and with legs parallel to those of the original triangle.

With the diagram explained, we can get to the argument-proper:

$$\begin{align} |QP|^2+|QK|^2=|KP|^2 &\quad\to\quad |a-r|^2 + |b-r|^2 = |c-r|^2 \\ &\quad\to\quad r^2 - 2 r ( a + b - c ) = c^2-a^2-b^2 = 0 \\ &\quad\to\quad r \;\left(\; r - 2 ( a + b - c ) \;\right) = 0 \\ &\quad\to\quad r = 2 ( a + b - c ) \end{align}$$

The reader can adjust the argument appropriately and find that the radius of the question's "external" circle is $2 ( a + b + c )$.

Thus, the product of the radii is $$2 ( a + b - c ) \cdot 2( a + b + c ) = 4 (( a + b )^2 - c^2 ) = 4 ((a+b)^2-(a^2+b^2)) = 8 a b$$ which is $4\;|\triangle ABC|$, as desired. $\square$

Answer 3

Solution

More generally,as the figure shows, for any $\triangle ABC$, let $a,b,c$ be the side length of $BC,CA,AB$ respectively. Then according to the properties of the mixtilinear circle, we have

$$AP=\frac{2bc}{b+c+a},~~~~AQ=\frac{2bc}{b+c-a}.$$

Now, let $A=90^{o}.$ Then $$x=AP,~~y=AQ,~~b^2+c^2=a^2,~~S_{\triangle ABC}=\frac{1}{2}bc.$$ It follows that$$xy=\frac{2bc}{b+c+a}\cdot \frac{2bc}{b+c-a}=\frac{4b^2c^2}{(b+c)^2-a^2}=\frac{4b^2c^2}{(b^2+c^2-a^2)+2bc}=2bc=4S_{\triangle ABC}$$