Triangle ABC (where ACB>90) is inscribed into circle with centre O. Line CO
intersects segment AB at point S. Prove that if AC+BC=2CO then radii of incircles of triangles ASC and BSC have the same length.
Could anyone help me with that please?
Triangle ABC (where ACB>90) is inscribed into circle with centre O. Line CO
intersects segment AB at point S. Prove that if AC+BC=2CO then radii of incircles of triangles ASC and BSC have the same length.
Could anyone help me with that please?
Hello matev91
Welcome to Math Help Forum!I have spent some time looking at this question, and have to report that it's not true - the radii of the incircles are not equal in length.
With the usual notation, the radius of the incircle of a triangle, , is given by the formulaand the radius of the circumcircle, , by the formulaThis second formula - the Sine Rule, of course - gives
and so if (i.e. ):I denoted the lengths of the line segments as follows:
and, as usual,and I also denoted the angle:Then, using the Sine Rule on :So the area of the triangle is given by:and hence the radius, , of its incircle (when simplified) by:In the same way, the radius, , of the incircle of is:
So the problem can now be re-stated as:Given that , prove that, for all valid values of ,It looks promising, but it just ain't so! The LHS can be manipulated as follows:
and, in a similar way, the RHS
So, in order for these to be equal, we would need
i.e. , for all values of , given . But this simply is not so.
I have put some calculations together into an Excel spreadsheet to confirm this - that we can fulfill all the conditions and yet produce different radii. I attach the Excel file.
Grandad
Hello again matev91
Indeed, I did get it wrong, and I now have the required proof - though it's still somewhat complicated. I wonder whether anyone can find a simpler one?
What I failed to spot was a simple relationship between the angles in the diagram, and it's this:
Produce the radius to meet the circle again at ; join . Then:(angle in a semicircle)
(angles in same segment)
Thus, with my original notation, (exterior angle of )
This gives us:
Now in my original posting I showed thatWe can now eliminate and (noting that , first from the LHS:
And the RHS:
And that completes the proof.
There has to be a simpler solution!
Grandad
PS Ah - I see I'm too late. Well done for finding the solution in spite of me! I feel sure you've got something simpler. Perhaps you'd like to post it here if so, so that everyone can see.
I also spent quite a lot of time on this problem. I came to the conclusion that it is correct, and I have a rather long and clumsy proof of it. However, I have seen the same problem posed in a number of online forums, including here, where it is claimed that it is a problem in the current Polish Math Olympiad, and hints and answers to it should not be given. It seems a shame to delete the posts here, when so much time and effort have been spent on them, but perhaps the Mods ought to do so nonetheless, if the information about the Polish Math Olympiad is correct (I have no idea how one might confirm that).