Categories
AMC-8 India Math Olympiad Math Olympiad PRMO USA Math Olympiad

Problem on Real numbers | Algebra | PRMO-2017 | Problem 18

Try this beautiful problem from Algebra based on real numbers from PRMO 2017. You may use sequential hints to solve the problem.

Try this beautiful problem from Algebra PRMO 2017 based on real numbers.

Problem on Real numbers | PRMO | Problem 18


If the real numbers \(x\), \(y\), \(z\) are such that \(x^2 + 4y^2 + 16z^2 = 48\) and \(xy + 4yz + 2zx = 24\). what is the
value of \(x^2 + y^2 + z^2\) ?

  • $24$
  • $21$
  • $34$

Key Concepts


Algebra

Equation

Check the Answer


But try the problem first…

Answer:\(21\)

Source
Suggested Reading

PRMO-2017, Problem 18

Pre College Mathematics

Try with Hints


First hint

The given equation are

\(x^2 + 4y^2 + 16z^2 = 48\)
\(\Rightarrow (x)2 + (2y)2 + (4z)2 = 48\)
\(2xy + 8yz + 4zx = 48\)
adding tis equations we have to solve the problem….

Can you now finish the problem ……….

Second Hint

Now we can say that
\((x)^2 + (2y)^2 + (4z)^2 – (2xy) – (8yz) – (4zx) = 0\)
\(\Rightarrow [(x – 2y)2 + (2y – 4z)2 + (x – 4y)2)] = 0\)
\(x = 2y = 4z \)

\(\Rightarrow \frac{x}{4}=\frac{y}{2}=z\)

Can you finish the problem……..

Final Step

Therefore we may say that,

\((x, y, z) = (4m, 2m, m)\)
\(x^2 + 4y^2 + 16z^2 = 48\)

\(16m^2 + 16m^2 + 16m^2 = 48\)
so \(m^2 = 1\)
\(x^2 + y^2 + z^2 = 21m^2 = 21\)

Subscribe to Cheenta at Youtube


Leave a Reply

This site uses Akismet to reduce spam. Learn how your comment data is processed.