Four More Incircles

Geometry Level 5

△ A B C \triangle ABC is the unit equilateral triangle. △ D E F \triangle DEF is an equilateral triangle. The four incircles are congruent. What is their radius? Express it as a − b c \sqrt \dfrac{a - \sqrt b}{c} , where b b is square-free, and submit a + b + c a+b+c

inspiritaion


The answer is 58.

Fletcher Mattox by Fletcher Mattox , 72, USA

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2 solutions

David Vreken
Feb 17, 2021

Extend C D CD to meet A B AB at G G , A E AE to meet B C BC at H H , B F BF to meet A C AC at I I , and let k = A G = B H k = AG = BH .

By the law of cosines on △ A B H \triangle ABH , A H = B H 2 + A B 2 − 2 ⋅ B H ⋅ A B ⋅ cos ⁡ ∠ A B H = k 2 + 1 − 2 ⋅ k ⋅ 1 2 = k 2 − k + 1 AH = \sqrt{BH^2 + AB^2 - 2 \cdot BH \cdot AB \cdot \cos \angle ABH} = \sqrt{k^2 + 1 - 2 \cdot k \cdot \frac{1}{2}} = \sqrt{k^2 - k + 1} .

Since △ A G D ∼ △ B H E \triangle AGD \sim \triangle BHE by AA similarity, A D = A G ⋅ A B A H = k ⋅ 1 k 2 − k + 1 = k k 2 − k + 1 = E B AD = AG \cdot \cfrac{AB}{AH} = k \cdot \cfrac{1}{\sqrt{k^2 - k + 1}} = \cfrac{k}{\sqrt{k^2 - k + 1}} = EB .

Also, G D = A F ⋅ B H A H = k ⋅ k k 2 − k + 1 = k 2 k 2 − k + 1 = E H GD = AF \cdot \cfrac{BH}{AH} = k \cdot \cfrac{k}{\sqrt{k^2 - k + 1}} = \cfrac{k^2}{\sqrt{k^2 - k + 1}} = EH .

Then D E = A H − A D − E H = k 2 − k + 1 − k k 2 − k + 1 − k 2 k 2 − k + 1 = 1 − 2 k k 2 − k + 1 DE = AH - AD - EH = \sqrt{k^2 - k + 1} - \cfrac{k}{\sqrt{k^2 - k + 1}} - \cfrac{k^2}{\sqrt{k^2 - k + 1}} = \cfrac{1 - 2k}{\sqrt{k^2 - k + 1}} .

And A E = A D + D E = k k 2 − k + 1 + 1 − 2 k k 2 − k + 1 = 1 − k k 2 − k + 1 AE = AD + DE = \cfrac{k}{\sqrt{k^2 - k + 1}} + \cfrac{1 - 2k}{\sqrt{k^2 - k + 1}} = \cfrac{1 - k}{\sqrt{k^2 - k + 1}} .

As an incircle of equilateral △ D E F \triangle DEF , r = 1 2 3 ⋅ D E = 1 2 3 ⋅ 1 − 2 k k 2 − k + 1 = 1 − 2 k 2 3 k 2 − k + 1 r = \cfrac{1}{2\sqrt{3}} \cdot DE = \cfrac{1}{2\sqrt{3}} \cdot \cfrac{1 - 2k}{\sqrt{k^2 - k + 1}} = \cfrac{1 - 2k}{2\sqrt{3}\sqrt{k^2 - k + 1}} .

As an incircle of △ A E B \triangle AEB , r = A E ⋅ E B ⋅ sin ⁡ ∠ A E B A E + E B + A B = 1 − k k 2 − k + 1 ⋅ k k 2 − k + 1 ⋅ 3 2 1 − k k 2 − k + 1 + k k 2 − k + 1 + 1 = 3 2 3 k 2 − k + 1 − 3 2 r = \cfrac{AE \cdot EB \cdot \sin \angle AEB}{AE + EB + AB} = \cfrac{\frac{1 - k}{\sqrt{k^2 - k + 1}} \cdot \frac{k}{\sqrt{k^2 - k + 1}} \cdot \frac{\sqrt{3}}{2}}{\frac{1 - k}{\sqrt{k^2 - k + 1}} + \frac{k}{\sqrt{k^2 - k + 1}} + 1} = \cfrac{3}{2\sqrt{3}\sqrt{k^2 - k + 1}} - \cfrac{\sqrt{3}}{2} .

So r = 1 − 2 k 2 3 k 2 − k + 1 = 3 2 3 k 2 − k + 1 − 3 2 r = \cfrac{1 - 2k}{2\sqrt{3}\sqrt{k^2 - k + 1}} = \cfrac{3}{2\sqrt{3}\sqrt{k^2 - k + 1}} - \cfrac{\sqrt{3}}{2} , which solves to k = 17 − 3 21 10 k = \cfrac{17 - 3\sqrt{21}}{10} and r = 5 − 21 32 r = \sqrt{\cfrac{5 - \sqrt{21}}{32}} .

Therefore, a = 5 a = 5 , b = 21 b = 21 , c = 32 c = 32 , and a + b + c = 58 a + b + c = \boxed{58} .

3 Helpful 2 Interesting 4 Brilliant 0 Confused

Nice - once again quite a different approach to mine. It's interesting how close the values of A D AD and D E DE are.

By the way, the answer can also be written as 7 − 3 8 \frac{\sqrt7-\sqrt3}{8}

Chris Lewis - 3 months, 3 weeks ago

Hi David, I always appreciate your solutions, which are amongst my favorites. Thanks so much. I have a question here : I do not understand the end of the first line. I would have expected √(k^2-k+1) instead of √(k^2-2k+1) as we have 2.k.(1/2).

Gerard Boileau - 3 months, 2 weeks ago

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Thanks, it should have been k 2 − k + 1 \sqrt{k^2 - k + 1} . It was a typo that I kept copying and pasting in the rest of the problem. It should be fixed now.

David Vreken - 3 months, 2 weeks ago

Thanks ! It's always a pleasure to see how you solve this kind of problem almost without trigonometry !

Gerard Boileau - 3 months, 2 weeks ago

Hi David, I fell really stupid but I cannot find out how to simplify the equation of ∆AEB incircle radius, to get r= 3/(2√3√(k^2-k+1) -√3/2. Would you be kind enough to give some intermediate results or tell "the trick" ? Thanks a lot.

Gerard Boileau - 3 months, 2 weeks ago

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If you start by multiplying the expression by k 2 − k + 1 k 2 − k + 1 \cfrac{\sqrt{k^2 - k + 1}}{\sqrt{k^2 - k + 1}} , then:

r = 1 − k k 2 − k + 1 ⋅ k k 2 − k + 1 ⋅ 3 2 1 − k k 2 − k + 1 + k k 2 − k + 1 + 1 ⋅ k 2 − k + 1 k 2 − k + 1 = ( 1 − k ) k k 2 − k + 1 ⋅ 3 2 1 + k 2 − k + 1 r = \cfrac{\frac{1 - k}{\sqrt{k^2 - k + 1}} \cdot \frac{k}{\sqrt{k^2 - k + 1}} \cdot \frac{\sqrt{3}}{2}}{\frac{1 - k}{\sqrt{k^2 - k + 1}} + \frac{k}{\sqrt{k^2 - k + 1}} + 1} \cdot \cfrac{\sqrt{k^2 - k + 1}}{\sqrt{k^2 - k + 1}} = \cfrac{\frac{(1 - k)k}{\sqrt{k^2 - k + 1}} \cdot \frac{\sqrt{3}}{2}}{1 + \sqrt{k^2 - k + 1}}

Then multiply the expression by 1 − k 2 − k + 1 1 − k 2 − k + 1 \cfrac{1 - \sqrt{k^2 - k + 1}}{1 - \sqrt{k^2 - k + 1}} , so:

r = ( 1 − k ) k k 2 − k + 1 ⋅ 3 2 ⋅ ( 1 − k 2 − k + 1 ) ( 1 + k 2 − k + 1 ) ( 1 − k 2 − k + 1 ) = ( 1 − k ) k k 2 − k + 1 ⋅ 3 2 ⋅ ( 1 − k 2 − k + 1 ) ( 1 − k ) k r = \cfrac{\frac{(1 - k)k}{\sqrt{k^2 - k + 1}} \cdot \frac{\sqrt{3}}{2} \cdot (1 - \sqrt{k^2 - k + 1})}{(1 + \sqrt{k^2 - k + 1})(1 - \sqrt{k^2 - k + 1})} = \cfrac{\frac{(1 - k)k}{\sqrt{k^2 - k + 1}} \cdot \frac{\sqrt{3}}{2} \cdot (1 - \sqrt{k^2 - k + 1})}{(1 - k)k}

Then simplifying:

r = ( 1 − k ) k k 2 − k + 1 ⋅ 3 2 ⋅ ( 1 − k 2 − k + 1 ) ( 1 − k ) k = 1 k 2 − k + 1 ⋅ 3 2 ⋅ ( 1 − k 2 − k + 1 ) = 3 2 3 k 2 − k + 1 − 3 2 r = \cfrac{\frac{(1 - k)k}{\sqrt{k^2 - k + 1}} \cdot \frac{\sqrt{3}}{2} \cdot (1 - \sqrt{k^2 - k + 1})}{(1 - k)k} = \cfrac{1}{\sqrt{k^2 - k + 1}} \cdot \cfrac{\sqrt{3}}{2} \cdot (1 - \sqrt{k^2 - k + 1}) = \cfrac{3}{2\sqrt{3}\sqrt{k^2 - k + 1}} - \cfrac{\sqrt{3}}{2} .

David Vreken - 3 months, 2 weeks ago

thank you so much ! I really do appreciate your help and how much I can learn from your elegant solutions ! Gerard.

Gerard Boileau - 3 months, 2 weeks ago

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You're very welcome!

David Vreken - 3 months, 2 weeks ago
Chew-Seong Cheong
Feb 18, 2021

Let the centers of the middle and the right circles be O O and P P respectively, O M OM and P N PN be perpendicular to A B AB , O K OK be perpendicular to A E AE , and O M OM and A E AE intersect at L L . Let ∠ E A B = ∠ L O K = θ \angle EAB = \angle LOK = \theta , then ∠ E B A = 6 0 ∘ − θ \angle EBA = 60^\circ - \theta , and the radius of the four congruent circles be r r .

Then we have:

A N + N B = A B r cot ⁡ θ 2 + r cot ⁡ ( 3 0 ∘ − θ 2 ) = 1 Let t = tan ⁡ θ 2 r t + 3 + t 1 − 3 t ⋅ r = 1 ⟹ r = t ( 1 − 3 t ) 1 + t 2 \begin{aligned} AN + NB & = AB \\ r \cot \frac \theta 2 + r \cot \left(30^\circ - \frac \theta 2\right) & = 1 & \small \blue{\text{Let }t = \tan \frac \theta 2} \\ \frac rt + \frac {\sqrt 3 + t}{1-\sqrt 3t} \cdot r & = 1 \\ \implies r & = \frac {t(1-\sqrt 3t)}{1+t^2} \end{aligned}

Note that O O is also the centroid of △ A B C \triangle ABC . Therefore O M = 3 6 OM = \frac {\sqrt 3}6 . Then

O L + L M = O M r sec ⁡ θ + 1 2 tan ⁡ θ = 3 6 1 + t 2 1 − t 2 ⋅ r + t 1 − t 2 = 3 6 ⟹ r = 3 − 6 t − 3 t 2 6 ( 1 + t 2 ) \begin{aligned} OL + LM & = OM \\ r \sec \theta + \frac 12 \tan \theta & = \frac {\sqrt 3}6 \\ \frac {1+t^2}{1-t^2} \cdot r + \frac t{1-t^2} & = \frac {\sqrt 3}6 \\ \implies r & = \frac {\sqrt 3 - 6t - \sqrt 3 t^2}{6(1+t^2)} \end{aligned}

Therefore

3 − 6 t − 3 t 2 6 ( 1 + t 2 ) = t ( 1 − 3 t ) 1 + t 2 5 3 t 2 − 12 t + 3 = 0 ⟹ t = 6 − 21 5 3 = 2 3 − 7 5 \begin{aligned} \frac {\sqrt 3 - 6t - \sqrt 3 t^2}{6(1+t^2)} & = \frac {t(1-\sqrt 3t)}{1+t^2} \\ 5\sqrt 3 t^2 - 12 t + \sqrt 3 & = 0 \\ \implies t & = \frac {6-\sqrt{21}}{5\sqrt 3} = \frac {2\sqrt 3 - \sqrt 7}5 \end{aligned}

Then r = t ( 1 − 3 t ) 1 + t 2 = 5 − 21 32 r = \dfrac {t(1-\sqrt 3t)}{1+t^2} = \sqrt{\dfrac {5-\sqrt{21}}{32}} . Therefore a + b + c = 58 a+b+c = \boxed{58} .

3 Helpful 2 Interesting 3 Brilliant 0 Confused

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