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$\Delta POR $ has vertices $P(0,12),R(5,0)$ and $O(0,0)$. There exists a line $l$ cutting $PR$ and $OP$ at $A$ and $B$ respectively such that circles can be inscribed in $\Delta PAB$ and quadrilateral $ORAB$. Also, these circles are tangent to the line $l$ at the same point. If line $l$ passes through the point $(0,8)$, the find the area of the quadrilateral $ORAB$.

I tried using the condition for a quadrilateral having an inscribed circle. I am not able to use the tangent at same point condition. Though I am not sure if my approach is the way to start. Any help is appreciated. Thanks.

Hoot
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  • Just a comment, I would say this probably falls more under the heading of analytic geometry, instead of algebraic geometry. – pjs36 Jul 10 '16 at 15:10
  • A little investment on drawing or illustration would have helped the cause.. – MonK Jul 11 '16 at 08:20
  • Is it really possible to draw such a figure meeting all the given conditions? – Mick Jul 11 '16 at 15:37

1 Answers1

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I am not able to use the tangent at same point condition.

This answer does not use the condition that the two circles are tangent to $l$ at the same point because $A$ is determined only by the condition that $ORAB$ possesses an incircle. (After knowing $A$, I see that the two circles are tangent to $l$ at $(18/5,16/5)$.)

The center and the radius of the circle inscribed in $ORAB$ can be written as $(r,r),r$ respectively where $r\lt 5$.

Since the distance between $PR: 12x+5y-60=0$ and $(r,r)$ is $r$, $$r=\frac{|12r+5r-60|}{\sqrt{12^2+5^2}}\implies r=2$$

Also, since the distance between $l:bx-y+8=0$ and $(2,2)$ is $r=2$, $$2=\frac{|2b-2+8|}{\sqrt{b^2+1}}\implies b=-\frac 43$$

Finally, $A$ is the intersection point of $l$ and $PR$, so $A(15/4,3)$ from which we get that the area of $ORAB$ is $\color{red}{45/2}$.

mathlove
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