Sum of red = Sum of blue?

Geometry Level 3

The figure below is made on the regular grid.

Which is bigger, the red angle, or the sum of the blue angles?

The red angle The sum of the blue angles They are equal

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1 solution

Tom Engelsman
Aug 6, 2017

Assuming each grid side length equals unity, we have:

θ r e d = a r c t a n ( 1 2 ) \theta_{red} = arctan(\frac{1}{2}) , θ b l u e 1 = a r c t a n ( 1 3 ) \theta_{blue1} = arctan(\frac{1}{3}) , θ b l u e 2 = a r c t a n ( 1 6 ) \theta_{blue2} = arctan(\frac{1}{6}) .

Knowing that the tangent function is an increasing function, let's now calculate:

t a n ( θ r e d ) = 1 2 . tan(\theta_{red}) = \frac{1}{2}.

t a n ( θ b l u e 1 + θ b l u e 2 ) = t a n ( θ b l u e 1 ) + t a n ( θ b l u e 2 ) 1 t a n ( θ b l u e 1 ) t a n ( θ b l u e 2 ) = 1 3 + 1 6 1 ( 1 3 ) ( 1 6 ) = 9 17 tan(\theta_{blue1} + \theta_{blue2}) = \frac{tan(\theta_{blue1}) + tan(\theta_{blue2})}{1 - tan(\theta_{blue1}) tan(\theta_{blue2})} = \frac{\frac{1}{3} + \frac{1}{6}}{1 - (\frac{1}{3})(\frac{1}{6})} = \frac{9}{17} .

Since 1 2 < 9 17 \frac{1}{2} < \frac{9}{17} , the blue angle sum is greater.

I suppose that last calculation should be tan ( θ b 1 + θ b 2 ) = 1 2 17 18 = 9 17 \tan(\theta_{b1} + \theta_{b2}) = \dfrac{\dfrac{1}{2}}{\dfrac{17}{18}} = \dfrac{9}{17} .

The final result is still the same as 1 2 < 9 17 \dfrac{1}{2} \lt \dfrac{9}{17} . :)

Brian Charlesworth - 3 years, 10 months ago

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Oops, my bad Brian.....all fixed!

tom engelsman - 3 years, 10 months ago

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