Categories
Geometry Math Olympiad PRMO USA Math Olympiad

Finding side of Triangle | PRMO-2014 | Problem 15

Try this beautiful problem from the Pre-Regional Mathematics Olympiad, PRMO, 2014, based on finding side of Triangle. You may use sequential hints.

Try this beautiful problem from PRMO, 2014 based on Finding side of Triangle.

Finding side of Triangle | PRMO | Problem 15


Let XOY be a triangle with angle XOY=90 degrees. Let M and N be the midpoints of the legs OX and OY, respectively. Suppose that XN=19 and YM=22. what is XY?

  • \(28\)
  • \(26\)
  • \(30\)

Key Concepts


Geometry

Triangle

Pythagoras

Check the Answer


Answer:\(26\)

PRMO-2014, Problem 15

Pre College Mathematics

Try with Hints


Finding side of Triangle - figure

Given that \(\angle XOY=90^{\circ}\) .Let M and N be the midpoints of the legs OX and OY, respectively, and that XN=19 and YM=22. Now \(\triangle XON\) & \(\triangle MOY\) are Right angle Triangle. Use Pythagoras theorem …….

Can you now finish the problem ……….

Triangle problem

Let \(XM=MO=p\) and \(ON=NY=q\).Now using Pythagoras theorm on \(\triangle XON\) & \(\triangle MOY\) we have…

\(OX^2 +ON^2=XN^2\) \(\Rightarrow 4p^2 +q^2=19^2\) \(\Rightarrow 4p^2 +q^2=361\)………..(1) and \(OM^2 +OY^2=MY^2\) \(\Rightarrow p^2 +4q^2=22^2\) \(\Rightarrow p^2 +4q^2=484\)……(2)

Finding side of Triangle

Now Adding (1)+(2)=\((4p^2 +q^2=361)\)+\((p^2 +4q^2=484\) \(\Rightarrow 5(p^2+q^2)=845\) \(\Rightarrow (p^2+q^2)=169\) \(\Rightarrow 4(p^2+q^2)=676\) \(\Rightarrow (OX)^2+(OY)^2=(26)^2\) \(\Rightarrow (XY)^2=(26)^2\) \(\Rightarrow XY=26\).

Subscribe to Cheenta at Youtube


Leave a Reply

Your email address will not be published. Required fields are marked *