Bored with congruent circles, yet?

Geometry Level 3

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 p q \frac{p}{q} , where p p and q q are coprime positive integers, submit p + q p +q .

Bonus: Generalise the solution for any right triangle in terms of the length of its legs.


The answer is 19.

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3 solutions

Chew-Seong Cheong
Oct 22, 2020

Consider a general right triangle with side lengths a a , b b , and c c . Let the radius of the four congruent circles be r r and C = θ \angle C = \theta . Also D E DE be the line passing through the centers right two circles, which is perpendicular to A C AC . Then D E = r sec θ + 3 r DE = r\sec \theta + 3r and E C = b 3 r EC = b-3r . Now we have:

D E E C = tan θ r sec θ + 3 r b 3 r = tan θ Note that sec θ = a b and tan θ = c b a b r + 3 r b 3 r = c b a r + 3 b r = b c 3 c r r = b c a + 3 ( b + c ) \begin{aligned} \frac {DE}{EC} & = \tan \theta \\ \frac {r\sec \theta + 3r}{b-3r} & = \tan \theta & \small \blue{\text{Note that }\sec \theta = \frac ab \text{ and }\tan \theta = \frac cb} \\ \frac {\frac ab r + 3r}{b-3r} & = \frac cb \\ ar +3br & = bc - 3cr \\ \implies r & = \frac {bc}{a+3(b+c)} \end{aligned}

For a = 5 a=5 , b = 4 b=4 , and c = 3 c=3 , r = 4 × 3 5 + 3 ( 4 + 3 ) = 12 26 = 6 13 r = \dfrac {4 \times 3}{5+3(4+3)} = \dfrac {12}{26} = \dfrac 6{13} . Then a + b = 6 + 13 = 19 a+b = 6+13 = \boxed{19} .

Nice job, Chew-Seong. As always. Thank you.

Fletcher Mattox - 7 months, 3 weeks ago

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You are welcome. For this problem and others when the answer is a b \dfrac ab , you have to mention positive coprime integers, because when both a a and b b are negative they are valid solution, so there will be two values for a + b a+b .

Chew-Seong Cheong - 7 months, 3 weeks ago
Toby M
Oct 22, 2020

No trigonometry is needed. The area of Δ A B C \Delta ABC can be written as Δ B G C + Δ B G D + Δ G F C + \Delta BGC + \Delta BGD + \Delta GFC + \ square A D G F ADGF , which can be expressed all in terms of r r :

1 2 ( 5 r + ( 3 3 r ) ( 3 r ) + ( 4 3 r ) ( 3 r ) ) + ( 3 r ) 2 = 12 2 \frac{1}{2} \left(5r + (3-3r)(3r) + (4-3r)(3r) \right) + (3r)^2 = \frac{12}{2} 5 r + 21 r 18 r 2 + 18 r 2 = 12 5r + 21r - 18r^2 + 18r^2 =12 r = 12 5 + 21 = 6 13 r = \frac{12}{5 + 21} = \frac{6}{13}

Hence p + q = 6 + 13 = 19 p + q = 6 + 13 = \boxed{19} .

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 = b c a + 3 ( b + c ) ar + 3cr + 3br = ab \Rightarrow r = \frac{bc}{a+3(b+c)} in general.

Actually if you connect A and G, then the equation becomes: 1 2 × ( 3 r × 4 + 3 r × 3 + r × 5 ) = 6 \dfrac{1}{2} \times\left( 3r \times 4+3r \times 3+r\times 5\right)=6

Isaac YIU Math Studio - 7 months, 3 weeks ago

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That's even better! From here, we get r = 12 26 = 6 13 r = \frac{12}{26} = \frac{6}{13} more directly.

Toby M - 7 months, 3 weeks ago

Exactly my first thought: why not divide that's square? But anyhow it's still a very nice solution!

Peter van der Linden - 7 months, 1 week ago

I think this is an ingeneous and creative solution @Toby M . Good work. I think you meant to say A B C \triangle{ABC} not A B C D \triangle{ABCD} .

Fletcher Mattox - 7 months, 3 weeks ago
Vinod Kumar
Nov 13, 2020

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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