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$\triangle ABC$ is right angle triangle and its circumcenter is $O$. $G$ is a point where $BC$ is tangent to the incircle. The perpendicular distance from $BC$ to circumcircle at $G$ is 10. How to calculate the area of $\triangle ABC$?

I have tried to prove if the incenter, circumcenter and orthocenter are collinear but failed. I couldn't find what was special about the point $G$. What would be the correct approach to solve this problem?

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  • $\begingroup$ Hint use the tangents . They are the key to the problem $\endgroup$ Aug 31, 2020 at 5:08
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    $\begingroup$ Hint: Join $BD$ and $DC$ to get $BG\times GC=100$ $\endgroup$
    – Anand
    Aug 31, 2020 at 5:15
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    $\begingroup$ @Anand is the area 100? $\endgroup$
    – Shromi
    Aug 31, 2020 at 5:44

2 Answers 2

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Let $|BC|=a$, $|AC|=b$, $|AB|=c$, $|GE|=|DE|$.

The distances to the tangent point $G$ of the incircle are

\begin{align} |BG|&=\tfrac12(a+c-b) \tag{1}\label{1} ,\\ |CG|&=\tfrac12(a+b-c) \tag{2}\label{2} , \end{align}

and by the power of the point $G$ w.r.t the circumcircle,

\begin{align} |BG|\cdot|CG|&=|DG|\cdot|EG|=|DG|^2=100 \tag{3}\label{3} ,\\ |BG|\cdot|CG|&=\tfrac14(a+c-b)(a+b-c) =\tfrac14(a^2-(c-b)^2) =\tfrac14(b^2+c^2-(c-b)^2) =\tfrac12\,bc \tag{4}\label{4} , \end{align}

hence, the area of $\triangle ABC$ is $100$.

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  • $\begingroup$ how did you get the lengths of BG and CG? $\endgroup$
    – C Squared
    Aug 31, 2020 at 7:51
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    $\begingroup$ @C Squared: It's a very well-known property of the tangent points of the incircle in any triangle, which can be easily deduced. Consider all three tangent points. Let the distances from the vertices $A,B,C$ to any adjacent tangent point be $x,y$ and $z$. Then $x+y=c,y+z=a,z+x=b$, find $x,y,z$. $\endgroup$
    – g.kov
    Aug 31, 2020 at 8:09
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Euler's theorem states that the distance d between the circumcentre and incentre of a triangle is given by $d^{2}=R(R-2r)$.

Let $I$ be a center of incircle. We have

$$ OI^2 = IG^2 + OG^2$$ or $$OG^2 = OI^2 - IG^2 = R(R-2r)-r^2.$$

On other hand, we have $$OD^2 = DG^2 + OG^2$$ or $$R^2 = DG^2 + (R^2-2Rr - r^2)$$ Then $$DG^2 = r(2R+r) = 100.$$

Note that $S_{ABC} = \frac{r(AB+BC+CA)}{2} = \frac{r(2r+4R)}{2}=r(2R+r).$

So we have $S_{ABC} = 100$.

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