Least Right Side

The sides of a right triangle are all positive integers. Two of the sides are odd numbers that differ by 50. What is the least possible value for the third side?
Source: mathcontest.olemiss.edu 2/18/2013

SOLUTION

By the Pythagorean theorem
a^2+b^2=c^2
a^2=c^2-b^2
=\left (c+b\right )\left (c-b\right )
=\left (c+b\right )\left (50\right )
The least perfect square number that is divisible by 50 is 3600
a^2=3600
a=60
Verification
3600=\left (c+b\right )50
c+b=72
c-b=50
Add the two equations
2c=122
c=61
b=61-50=11
60^2+11^2=61^2

Answer: 60

Advertisements

About mvtrinh

Retired high school math teacher.
This entry was posted in Problem solving and tagged , , , , . Bookmark the permalink.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s