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Puzzles
Arctan
Source:
Futility Closet
Prove that \(\arctan(1)+\arctan(2)+\arctan(3)=\pi\).
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Let \(\alpha=\arctan(1)\), \(\beta=\arctan(2)\) and \(\gamma=\arctan(3)\), then draw the angles as follows:
Then proceed as in
Three Squares
.
Extension
Can you find any other integers \(a\), \(b\) and \(c\) such that:
$$\arctan(a)+\arctan(b)+\arctan(c)=\pi$$
Tags:
geometry
,
2d shapes
,
triangles
,
trigonometry
If you enjoyed this puzzle, check out
Sunday Afternoon Maths XXXVIII
,
puzzles about
trigonometry
, or
a random puzzle
.
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List of all puzzles
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irreducible numbers
star numbers
mean
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quadrilaterals
area
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ellipses
factors
sets
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sums
chess
complex numbers
numbers grids
pascal's triangle
dates
functions
tiling
neighbours
planes
surds
chalkdust crossnumber
shapes
integration
proportion
hexagons
menace
determinants
people maths
cube numbers
elections
sport
algebra
binary
triangles
triangle numbers
gerrymandering
means
tournaments
chocolate
2d shapes
the only crossnumber
taxicab geometry
range
rugby
squares
time
multiples
partitions
sum to infinity
books
sequences
expansions
multiplication
division
consecutive integers
volume
bases
geometric mean
parabolas
cryptic clues
palindromes
matrices
floors
numbers
odd numbers
cryptic crossnumbers
folding tube maps
arrows
crosswords
money
products
addition
remainders
routes
lines
median
graphs
angles
advent
dodecagons
prime numbers
cubics
dominos
number
geometric means
trigonometry
probability
spheres
crossnumbers
grids
logic
albgebra
digital products
cards
dice
wordplay
polygons
colouring
games
probabilty
powers
averages
geometry
unit fractions
tangents
square grids
integers
coordinates
regular shapes
medians
digital clocks
square numbers
indices
axes
differentiation
ave
perimeter
circles
fractions
rectangles
quadratics
factorials
scales
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shape
even numbers
speed
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consecutive numbers
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calculus
christmas
percentages
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symmetry
balancing
decahedra
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