# circle inscribed into isosceles triangle

i am trying to solve following problem: suppose that legs AB=BC=30 in isosceles triangle,and center of inscribed circle divides altitude into 12:5 part,our aim is to find base,my problem is that i dont know what is equal radius or there is know any angle,so could not understand how to solve it,even i could not use Pythagorean theorem,because it would take a long calculations,please help me guys,i am trying to solve it right now,but no success yet

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isosceles circle ? –  pedja Apr 4 '12 at 13:19
sorry i have edit it –  dato datuashvili Apr 4 '12 at 13:21

Let $O$ be the centre of the incircle. Join $O$ to the three vertices.

It is not absolutely clear what ratio $12:5$ means. Let $M$ be the middle of the base. Maybe (i) $BO:OM=12:5$, or maybe (ii) $BO:OM=5:12$. (It will turn out that (ii) can't happen.)

We examine the consequences of (i). Let the base be $2y$ (I don't like fractions). Let $BO=12t$ and let $OM=5t$.

We calculate the area of triangle $BAC$ in two different ways. The area is half of base times height, so it is $(1/2)(2y)(17t)$, that is, $17yt$.

The area is also the sum of the areas of $\triangle BOA$, $\triangle BOC$, and $\triangle OAC$. This sum is $(1/2)(30)(5t)+(1/2)(30)(5t)+(1/2)(2y)(5t)$, which is $150t +5yt$. We conclude that $$17yt=150t+5yt.$$ Thus $12y=150$, and therefore $2y=25$.

Next we examine possibility (ii). The analysis is similar. We find pretty quickly that $y$ is "too big" (the Triangle Inequality is violated).

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@dato: If we let the altitude from B through O meet the base at D, I was reading the question to be $BO=\frac{12}5 OD$. dato: am I right? –  Ross Millikan Apr 4 '12 at 13:47
yes yes i understood now,thanks guys –  dato datuashvili Apr 4 '12 at 13:50

Hint :

$\Delta OBD \sim \Delta AB'B$

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Rather $\Delta ODB \sim \Delta AB'B$ –  Marc van Leeuwen Apr 4 '12 at 14:19