Given a 3-4-5 right triangle, find the radius of one of the four congruent circles as shown in the figure. If the radius is expressed as q p , where p and q are coprime positive integers, submit p + q .
Bonus: Generalise the solution for any right triangle in terms of the length of its legs.
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Nice job, Chew-Seong. As always. Thank you.
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You are welcome. For this problem and others when the answer is b a , you have to mention positive coprime integers, because when both a and b are negative they are valid solution, so there will be two values for a + b .
No trigonometry is needed. The area of Δ A B C can be written as Δ B G C + Δ B G D + Δ G F C + square A D G F , which can be expressed all in terms of r :
2 1 ( 5 r + ( 3 − 3 r ) ( 3 r ) + ( 4 − 3 r ) ( 3 r ) ) + ( 3 r ) 2 = 2 1 2 5 r + 2 1 r − 1 8 r 2 + 1 8 r 2 = 1 2 r = 5 + 2 1 1 2 = 1 3 6
Hence p + q = 6 + 1 3 = 1 9 .
Labelling the triangle's sides the same as Chew-Seong does, we get that a r + 3 c r + 3 b r = a b ⇒ r = a + 3 ( b + c ) b c in general.
Actually if you connect A and G, then the equation becomes: 2 1 × ( 3 r × 4 + 3 r × 3 + r × 5 ) = 6
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That's even better! From here, we get r = 2 6 1 2 = 1 3 6 more directly.
Exactly my first thought: why not divide that's square? But anyhow it's still a very nice solution!
I think this is an ingeneous and creative solution @Toby M . Good work. I think you meant to say △ A B C not △ A B C D .
There are two solutions r= 6/13 and r=3/4 to the equation ✓[(3-3r)^2+(3r)^2-r^2]+ ✓[(4-3r)^2+(3r)^2-r^2]=5.
Answer = 19
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Consider a general right triangle with side lengths a , b , and c . Let the radius of the four congruent circles be r and ∠ C = θ . Also D E be the line passing through the centers right two circles, which is perpendicular to A C . Then D E = r sec θ + 3 r and E C = b − 3 r . Now we have:
E C D E b − 3 r r sec θ + 3 r b − 3 r b a r + 3 r a r + 3 b r ⟹ r = tan θ = tan θ = b c = b c − 3 c r = a + 3 ( b + c ) b c Note that sec θ = b a and tan θ = b c
For a = 5 , b = 4 , and c = 3 , r = 5 + 3 ( 4 + 3 ) 4 × 3 = 2 6 1 2 = 1 3 6 . Then a + b = 6 + 1 3 = 1 9 .