Arithmetic progression, a cool stuff

Geometry Level 3

if length of sides of a right angled triangle are in A.P, then find the ratio of sides opposite to the acute angle

4:5 3:4 1:2 none of these 2:3

Sanketh Yelamanchili by sanketh Yelamanchili , 19, India

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1 solution

Marta Reece
Jun 11, 2017

If the arithmetic progression is a , a + d , a + 2 d a, a+d, a+2d , then Pythagorean theorem will take the form a 2 + ( a + d ) 2 = ( a + 2 d ) 2 a^2+(a+d)^2=(a+2d)^2

Solving for d d will produce d = a ( − 1 + 2 3 ) = a 3 d=a\left(\dfrac{-1+2}{3}\right)=\dfrac a3

Second solution, namely d = a ( − 1 − 2 3 ) = − a d=a\left(\dfrac{-1-2}{3}\right)=-a , does not correspond to a triangle as the progression is a , 0 , − a a, 0, -a

The ratio of the two shorter sides for the only good solution is then a a + a 3 = 3 4 = 3 : 4 \dfrac{a}{a+\frac a3}=\dfrac34=\boxed{3:4}

The triangle could have a = 3 a=3 and then it would have sides 3 , 4 , 5 3, 4, 5 .

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There is a typo in the wording of the problem. It should say "acute angles" not "acute angle."

Also, the ratio is not actually defined, as there are two angles and either one of them could be first. It would be better if it stated which should come first in some form.

Marta Reece - 4 years ago

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