Black Book 5

Geometry Level 5

Let the incircle of Δ A B C \Delta ABC touches the sides B C , C A , A B BC,CA,AB at A 1 , B 1 , C 1 {A}_{1},{B}_{1},{C}_{1} respectively.The incircle of Δ A 1 B 1 C 1 \Delta{A}_{1}{B}_{1}{C}_{1} touches its sides of B 1 C 1 , C 1 A 1 , A 1 B 1 {B}_{1}{C}_{1},{C}_{1}{A}_{1},{A}_{1}{B}_{1} at A 2 , B 2 , C 2 {A}_{2},{B}_{2},{C}_{2} respectively and so on.

  • If lim ⁡ n → ∞ ∠ A n = p \displaystyle \lim _{ n\rightarrow \infty }{ \angle A_{ n } } = { p }

  • And In Δ A 4 B 4 C 4 \Delta {A}_{4}{B}_{4}{C}_{4} , the value of ∠ A 4 \angle A_4 is q {q} .

Then p + q {p+q} can be expressed in its simplest form as m π + n A r \dfrac{{m}\pi+{n}A}{{r}} , where m , n m,n and r r are integers.

What is m + n + r {m+n+r} ?


The answer is 82.

Mehul Chaturvedi by Mehul Chaturvedi , 22, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Mehul Chaturvedi
Mar 6, 2016

  • It can clearly be seen by geometry the ∠ A 1 = 180 − ( B + C 2 ) \angle A_1 =180-(\dfrac{B+C}{2}) i.e 90 − ∠ A 2 90-\dfrac{\angle A}{2}

So we can generalize ∠ A n = ∠ A n − 1 2 \angle A_n= \dfrac{\angle A_{n-1}}{2}

A 1 = 90 − A 2 A 2 = 90 − 90 − A 2 2 A ∞ = 90 − 90 − 90 − 90 − 90 − … … … 2 2 2 2 A ∞ = 90 − A ∞ 2 A ∞ = 6 0 ∘ = π 3 { A }_{ 1 }{ = }{ 90 }-\dfrac { A }{ 2 } \\ { A }_{ 2 }{ = }{ 90 }-\dfrac { { 90 }-\dfrac { A }{ 2 } }{ 2 } \\ { A }_{ \infty }{ = }{ 90 }-\dfrac { 90-\dfrac { 90-\dfrac { 90-\dfrac { 90-\dots \dots \dots }{ 2 } }{ 2 } }{ 2 } }{ 2 } \\ { A }_{ \infty }{ = }{ 90 }-\dfrac { { A }_{ \infty } }{ 2 } \\ { A }_{ \infty }=60^{\circ}=\color{#D61F06}{\dfrac{\pi}{3}}

  • Now by the result ∠ A n = ∠ A n − 1 2 \angle A_n= \dfrac{\angle A_{n-1}}{2} we can easily get A 4 = 5 π + A 16 A_4=\color{#3D99F6}{\dfrac{5 \pi +A}{16}}

∴ p + q = 31 π + 3 A 48 \therefore p+q=\dfrac{31 \pi+3A}{48}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Seems you are enjoying your holidays with brilliant, BTW nice question.

Shubhendra Singh - 5 years, 3 months ago

Log in to reply

yeah thats true

Mehul Chaturvedi - 5 years, 3 months ago

If you solve that recurrence relation, you get A n = π 3 + ( − 1 2 ) n ( A − π 3 ) A_n=\frac{\pi}{3}+\left(\frac{-1}{2}\right)^n\left(A-\frac{\pi}{3}\right)

Now take n → ∞ n\rightarrow\infty and n = 4 n=4 .

A Former Brilliant Member - 5 years, 2 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...