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

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


Answer:\(21\)

PRMO-2017, Problem 18

Pre College Mathematics

Try with Hints


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 ……….

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……..

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

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AMC 8 India Math Olympiad Math Olympiad PRMO USA Math Olympiad

Roots of Equations | PRMO-2016 | Problem 8

Try this beautiful problem from Algebra based on roots of equations.

Roots of Equations | PRMO | Problem 8


Suppose that \(a\) and \(b\) are real numbers such that \(ab \neq 1\) and the equations \(120 a^2 -120a+1=0\) and \(b^2-120b+120=0\) hold. Find the value of \(\frac{1+b+ab}{a}\)

  • $200$
  • $240$
  • $300$

Key Concepts


Algebra

quadratic equation

Roots

Check the Answer


Answer:$240$

PRMO-2016, Problem 8

Pre College Mathematics

Try with Hints


The given equations are \(120 a^2 -120a+1=0\) and \(b^2-120b+120=0\).we have to find out the values of \(a\) and \(b\)….

Let \(x,y\) be the roots of the equation \(120 a^2 -120a+1=0\)then \(\frac{1}{x},\frac{1}{y}\) be the roots of the equations of \(b^2-120b+120=0\).can you find out the value of \(a\) & \(b\)

Can you now finish the problem ……….

From two equations after sim[lificatiopn we get…\(a=x\) and \(b=\frac{1}{y}\) (as \(ab \neq 1)\)

Can you finish the problem……..

\(\frac{1+b+ab}{a}\)=\(\frac { 1+\frac{1}{y} +\frac{x}{y}}{x}\)=\(\frac{(x+y)+1}{xy}=240\)

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AMC 8 Math Olympiad USA Math Olympiad

Time and Work | PRMO-2017 | Problem 3

Try this beautiful problem from PRMO, 2017 based on Time and work.

Time and work | PRMO | Problem-3


A contractor has two teams of workers : team A and team B. Team A can complete a job in 12 days and team B can do the same job in 36 days. Team A starts working on the job and team B joins team A after four days. The team A withdraws after two more days. For how many more days should team B work to complete the job ?

  • $20$
  • $16$
  • $13$

Key Concepts


Arithmetic

multiplication

unitary method

Check the Answer


Answer:$16$

PRMO-2017, Problem 3

Pre College Mathematics

Try with Hints


In the problem,we notice that first 4 days only A did the work.so we have to find out A’s first 4 days work done.next 2 days (A+B) did the work together,so we have to find out (A+B)’s 2 days work.

so we may take the total work =1

A’s 1 day’s work= \(\frac{1}{12}\) and B’s 1 day’s work=\(\frac{1}{36}\)

Can you now finish the problem ……….

Now B did complete the remaining work.so you have to find out the remaining work and find out how many more days taken….

so to find the remaining work subtract (A’s 4 day;s work + (A+B)’S 2 days work)) from the total work

Can you finish the problem……..

Let the total work be 1

A can complete the total work in 12 days,so A’S 1 day’s work=\(\frac{1}{12}\)

B can complete the total work in 36 days, so B’s 1 day’s work=\(\frac{1}{36}\)

First 4 days A’s workdone=\(\frac{4}{12}=\frac{1}{3}\)

After 4 days B joined and do the work with A 2 days

So \((A+B)\)’s 2 day’s workdone=\(2 \times( \frac{1}{12}+\frac{1}{36})\)=\(\frac{2}{9}\)

Remaining workdone=\((1-\frac{1}{3}-\frac{2}{9}\))=\(\frac{4}{9}\)

B will take the time to complete the Remaining work=\(36 \times \frac{4}{9}\)=16

Hence more time taken=16

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AMC 10 AMC 12 AMC 8 India Math Olympiad Math Olympiad PRMO Regional Mathematics Olympiads USA Math Olympiad

Can we prove that the length of any side of a triangle is not more than half of its perimeter?

Can we Prove that ……..


The length of any side of a triangle is not more than half of its perimeter

Key Concepts


Triangle Inequality

Perimeter

Geometry

Check the Answer


Answer: Yes we can definitely prove that by Triangle Inequality

Mathematical Circles – Chapter 6 – Inequalities Problem 3

Mathematical Circles by Dmitri Fomin , Sergey Genkin , Llia Itenberg

Try with Hints


We can start this sum by using this picture below

The length of the three sides of this triangle are a,b and c. So if we apply triangle inequality which implies that the length of one side of a triangle is less than the sum of the lengths of the two sides of that triangle. In reference to the theorem

b + c > a

Proof based on triangle

So can you try to do the rest of the sum ????????

According to the question we have to find the perimeter at first

Perimeter is the sum of the length of all sides of the triangle = a + b + c

And the length of each side is a or b or c.

We have to prove : a + b + c > length of any one side

This can be one of the most important hint for this problem. Try to do the rest of the sum …………………………..

Here is the rest of the sum :

As stated above if we use triangle inequality :

b + c > a

Lets add a to both the sides

a + b + c > a + a

a + b + c > 2 a

The left hand side of the above inequality is the perimeter of this triangle.

perimeter > 2 a

So , \(\frac {perimeter}{2} > a \)

\(\frac {perimeter}{2} \) = semi perimeter

Hence this is proved that the length of one side of a triangle is less than half of its perimeter.

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