Can I use a calculator?

Geometry Level 3

A = tan 1 ( 1 2 ) + tan 1 ( 1 3 ) + tan 1 ( 1 4 ) B = π 2 \begin{aligned} A &=& \tan^{-1} \left( \frac 1 {\sqrt2} \right)+ \tan^{-1} \left( \frac 1 {\sqrt3} \right) + \tan^{-1} \left( \frac 1 {\sqrt4} \right) \\ \\ B & = & \frac {\pi}{2} \\ \end{aligned}

Without resorting to tables or calculator, determine which of these numbers above is larger.

All credit for this problem goes to Pi Han Goh .

A A They are equal B B Not enough information

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3 solutions

Trevor Arashiro
May 10, 2015

Solution credit to Pi Han Goh

This question was inspired by Spiral of Theodorus .

Draw the first four triangles as shown above. The total angle subtended from the first triangle to the fourth triangle is n = 1 4 tan 1 ( 1 n ) \displaystyle \sum_{n=1}^4 \tan^{-1} \left(\frac 1 {\sqrt n} \right ) .

Using a protractor, we can see that it's larger than 13 5 = 3 π 4 135^\circ = \frac{3\pi}{4} .

Then, n = 1 4 tan 1 ( 1 n ) > 3 π 4 \displaystyle \sum_{n=1}^4 \tan^{-1} \left(\frac 1 {\sqrt n} \right ) > \frac{3\pi}{4} .

Because tan 1 ( 1 1 ) = π 4 \tan^{-1} \left ( \frac {1}{\sqrt 1} \right) = \frac {\pi}{4} , then n = 2 4 tan 1 ( 1 n ) > π 2 \displaystyle \sum_{n=2}^4 \tan^{-1} \left(\frac 1 {\sqrt n} \right ) > \frac{\pi}{2} or simply A > B A > B .

My last comment, if it posted (my phone is acting really weird lately, sorry), should say 2 tan 1 ( ( 1 3 ) 2\tan^{-1}(\left(\frac{1}{\sqrt3}\right) . And also, the first sentence should state that it's increasing at a decreasing rate. but, the difference between 1/2 and 1 3 \frac{1}{\sqrt{3}} is much smaller than that of 1 3 \frac{1}{\sqrt{3}} and 1 2 \frac{1}{\sqrt{2}}

Trevor Arashiro - 5 years, 9 months ago

It is not that much greater than 135 degrees, so ideally there should be a better solution.

Calvin Lin Staff - 5 years, 8 months ago

It is not greater than 135 degrees it is just a little over 90 degrees

Aditya Chauhan - 5 years, 8 months ago

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He added tan 1 1 = 4 5 \tan^{-1} 1 = 45 ^ \circ to the count. Look at the image.

Calvin Lin Staff - 5 years, 8 months ago

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Got it. Thanks!

Aditya Chauhan - 5 years, 8 months ago
Sayantan Mondal
Mar 5, 2018

Just it's way bigger...😂😂😂

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