enter image description here

$\triangle ABC$ 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?

  • $\begingroup$ Hint use the tangents . They are the key to the problem $\endgroup$ Aug 31, 2020 at 5:08
  • 2
    $\begingroup$ Hint: Join $BD$ and $DC$ to get $BG\times GC=100$ $\endgroup$
    – Anand
    Aug 31, 2020 at 5:15
  • 1
    $\begingroup$ @Anand is the area 100? $\endgroup$
    – Shromi
    Aug 31, 2020 at 5:44

2 Answers 2


enter image description here

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 $100$.

  • $\begingroup$ how did you get the lengths of BG and CG? $\endgroup$
    – C Squared
    Aug 31, 2020 at 7:51
  • 1
    $\begingroup$ @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

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 $S_{ABC} = 100$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.