SOAB)+S(OAC)=2S(OBC)

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SOAB)+S(OAC)=2S(OBC)

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power of a point

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Source: Iran 2005

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#1 h Aug 27, 2005, 1:03 PM • 4 Y

Y by Adventure10, Mango247, Mango247, Mango247

Suppose $O$ is circumcenter of triangle $ABC$ . Suppose $\frac{S(OAB)+S(OAC)}2=S(OBC)$ . Prove that the distance of $O$ (circumcenter) from the radical axis of the circumcircle and the 9-point circle is $\frac {a^2}{\sqrt{9R^2-(a^2+b^2+c^2)}}$

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#2 h Aug 27, 2005, 3:04 PM • 3 Y

Y by Adventure10, Adventure10, Mango247

I suppose the notation $S(OAB)$ is the same as $S_{\triangle OAB}$ or $[OAB]$

I haven't gave a thought in it, but I believe it can be solved by using the formula $OH^2=9R^2-(a^2+b^2+c^2)$ .

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Amir.S

#3 h Aug 27, 2005, 4:08 PM • 2 Y

Y by Adventure10, Mango247

$S_{ABC}=3S_{BOC}(1)$ let $AO$ intersect $BC$ at $A'$ from $(1)$ we can get $\frac {AO}{OA'}=2$ so Euler line is parallel to $BC$ .
in triangle $AOH$ use pythagores theorem and $OH^2=9R^2-(a^2+b^2+c^2)$ formula .
with Casi theorem we can calculate the distancse

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#4 h Sep 3, 2005, 7:48 AM • 2 Y

Y by Adventure10, Mango247

What does Casi theorem say?

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#5 h Sep 3, 2005, 9:50 AM • 2 Y

Y by Adventure10, Mango247

If it is Casey Theorem, then you can find it in the Geometry Formula section.

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jred

#6 h Aug 26, 2016, 2:14 PM • 2 Y

Y by Adventure10, Mango247

the statement is wrong, if $\triangle ABC$ is acute, then the distance should be $\frac {a^2}{4\sqrt{9R^2-(a^2+b^2+c^2)}}$ . and I believe that the result will be different when $\triangle ABC$ is obtuse.

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