
#1
Feb2912, 08:20 PM

P: 18

1. The problem statement, all variables and given/known data
Please see picture attached. 2. Relevant equations 3. The attempt at a solution If I use the pythagoras formula with the ratio I get the height as 3 yet it is longer than the side that is 4. Am I doing something wrong? Thanks. 



#2
Feb2912, 08:50 PM

Mentor
P: 21,071





#3
Feb2912, 08:54 PM

P: 19

The problem is that the illustration is not drawn to scale. Side BC should be shorter than AC, as can be seen by the ratio. What to do next, I don't know. 



#4
Feb2912, 09:04 PM

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P: 21,071

Height of triangle 



#5
Feb2912, 09:13 PM

P: 19





#6
Feb2912, 09:51 PM

P: 19

I believe I've come up with something.
The height in terms of the radius is going to be 2r for the portion of the height enclosed by the circle plus some remaining height x. If we solve for the ratio of the height to the other sides as mentioned previously, we get that the height is h, the side DC is (4/3)h and the side BC is (5/3)h. We know that the angles caused by the radius intersecting the side BC are each 90 degrees, because the side is tangent to the circle. We also know that the angle between the height and side DC is 90 degrees. Therefore length b (in attachment) is equal to side DC, which is (4/3)h. We know that the total length of side BC is (5/3)h, so length a (in attachment) is therefore (1/3)h. We know by Pythagorean's theorem, then, that a^2 + r^2 = (hr)^2. If we substitute (1/3)h for a and then substitute (2r + x) for h, we get ((1/3)(2r+x))^2+r^2 = (x+r)^2. You should be able to solve for x and plug it into h = 2r + x to get the height in terms of r. I have never worked with inscribed circles before, so there might be an error in this. That is the way I would go about it, though. Note: attachment is NOT drawn to scale, whatsoever! 



#7
Mar112, 12:43 AM

P: 18

thank you :)




#8
Mar112, 03:07 AM

P: 339

I have a solution but it's not a very elegant one, nor is it even satisfactory. See attachment to refer to the variables. After naming the variables in the pic, you can write the following three equations.
s^{2}  6s + h^{2}  8h  p^{2} + 10p = 0 s^{2} + h^{2}  p^{2} = 0 d^{2} = (3  s  d)^{2}  h^{2} 



#9
Mar112, 09:56 AM

P: 21

I have yet another idea of how to approach this:
All lines from the circles centre is going to bisect each angle at the triangle's corners. So lets add one more point to the traingle where that extra line, r is touching line BC  we'll call that point E. Line OE is hitting BC at a right angle. Line OC bisects the angle BCA. Due to the ration of sides given we know that BC = 5x, CD = 4x, and BD = sqrt of (5x)^2  (3x)^2 = 4x Now let's find some similar triangles. We can see that triangle OCD is the same as OCE so therefore CD =CE = 4x This means that the length of line BE is going to be 5x4x = 1x Looking at triangle OBE we want to figure out the length of line OB (we already know that OD is = to r) The length of line OB is equal to r + some value we will call Z So looking at triangle OBE we see that: (x)^2 + (r)^2 = (r+Z)^2 So Z = sqrt((x)^2 + (r)^2)  r Substitute x for , say CD/4, we get the length of BD as: BD = sqrt((CD/4)^2 + (r)^2)  r + 2r BD = sqrt((CD/4)^2 + (r)^2) + r 


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