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Hi! This problem looks cool but it's giving me a little trouble. Could someone help?
Thanks!!
In triangle ABC, AB = 10, AC = 8, and BC = 6. Let P be the point on the circumcircle of triangle ABC so that angle PCA = 45. Find CP.
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Hi;
This problem is easily solved with analytical geometry, this diagram shows you what you need to know:
I am getting
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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hi mathattack
This looks like it ought to have a simple solution but it is eluding me too. I can only see a solution involving the cosine rule for this. Basically this:
With those lengths ACB = 90, so AB is a diameter, and AOP = 90 too (angle at centre = 2x angle at edge )
So you can calculate AP and then use the cosine rule on triangle APC to get PC.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Thanks for your help Bobbym and Bob!
Both suggestions got me going and helped me to confirm my answer too. I was able to get it with similar triangles in the end, a la power of a point:
If you label the point where CP crosses AB and call it D, then
6x5⍻2 = CDxDP = BD(10-BD), also 5⍻2xCD = 8xBD
I ended up with BD=30/7, CD = 24⍻2/7 & DP = 25⍻2/7, which matched Bobbym's 7⍻2 for CP.
You both are great by the way. I've been following the forum for a while now, but only recently thought to add some posts of my own. Till next time.
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You are half right. Bob is a really good mathematician, I just hang here and try to look smart.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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