$\log_2 (3)+\log_6 (8)$

A. Their sum is less than zero.

B. Their sum is greater than zero.

Answer:

In order to answer the question correctly, we must be super familiar with how the graph of $y=\log_{10} x$ behaves.

A collection of intriguing competition level problems for secondary school students.

$\log_2 (3)+\log_6 (8)$

A. Their sum is less than zero.

B. Their sum is greater than zero.

Answer:

In order to answer the question correctly, we must be super familiar with how the graph of $y=\log_{10} x$ behaves.

Prove $\dfrac{\pi}{4}+\dfrac{1}{6}\gt \arctan\left({\dfrac{6}{5}}\right)$.

Do you know if we're pretty familiar with how the graph of a particular function behaves on certain interval, we could set up a definite integral to prove some of the inequality problems (be them hard, moderately hard or very difficult)?

Do you know if we're pretty familiar with how the graph of a particular function behaves on certain interval, we could set up a definite integral to prove some of the inequality problems (be them hard, moderately hard or very difficult)?

Prove $\dfrac{\pi}{4}+\dfrac{1}{6}\gt \arctan\left({\dfrac{6}{5}}\right)$.

Let's pick up where we left off...we needed to complete the proof using the readily available formula that says

$\arctan\left({\dfrac{6}{5}}\right)=\dfrac{\pi}{4}+\dfrac{1}{10}-\dfrac{1}{100}+\dfrac{1}{1500}-\dfrac{1}{125000}+\dfrac{1}{750000}-\dfrac{1}{8750000}+\cdots$

such that we get

Let's pick up where we left off...we needed to complete the proof using the readily available formula that says

$\arctan\left({\dfrac{6}{5}}\right)=\dfrac{\pi}{4}+\dfrac{1}{10}-\dfrac{1}{100}+\dfrac{1}{1500}-\dfrac{1}{125000}+\dfrac{1}{750000}-\dfrac{1}{8750000}+\cdots$

such that we get

What methods one could use to prove inequalities IMO problems?

Off the top of your head, you might want to shout out that AM-GM inequality, Cauchy Schwarz inequality. Jensen's inequality are among the "hot" and popular methods that you would consider using to effectively prove the inequality hard IMO problem.

Off the top of your head, you might want to shout out that AM-GM inequality, Cauchy Schwarz inequality. Jensen's inequality are among the "hot" and popular methods that you would consider using to effectively prove the inequality hard IMO problem.

Prove that $x^7-2x^5+10x^2-1$ has no root greater than 1.

Heuristic method:

The trick is to substitute $y+1$ for $x$ in the given function:

If $f(x)=x^7-2x^5+10x^2-1$, then after the substitution we see that we have:

Heuristic method:

The trick is to substitute $y+1$ for $x$ in the given function:

If $f(x)=x^7-2x^5+10x^2-1$, then after the substitution we see that we have:

Prove that $x^7-2x^5+10x^2-1$ has no root greater than 1.

In the previous blog post, I mentioned of solving the equation $x^7-2x^5+10x^2-1=0$ so to show that the function $x^7-2x^5+10x^2-1$ has no root greater than 1.

But, that is a really bad idea. The reason why the question setters stated the problem so mostly because they wanted to avoid us to solve for the problem. To solve for the polynomial of degree seven is really difficult, plus, the polynomial couldn't be factorized and so the real roots are kind of "ugly", there are no exact values for them.

In the previous blog post, I mentioned of solving the equation $x^7-2x^5+10x^2-1=0$ so to show that the function $x^7-2x^5+10x^2-1$ has no root greater than 1.

But, that is a really bad idea. The reason why the question setters stated the problem so mostly because they wanted to avoid us to solve for the problem. To solve for the polynomial of degree seven is really difficult, plus, the polynomial couldn't be factorized and so the real roots are kind of "ugly", there are no exact values for them.

Prove that $x^7-2x^5+10x^2-1$ has no root greater than 1.

This is one thought provoking problem, as it presents the golden opportunity for one to think hard and long so they understand the problem more well and that eventually will improve one's reasoning skill.

This is one thought provoking problem, as it presents the golden opportunity for one to think hard and long so they understand the problem more well and that eventually will improve one's reasoning skill.

Prove $35\sqrt{55}+55\sqrt{77}+77\sqrt{35}+35\sqrt{77}+55\sqrt{35}+77\sqrt{55}\gt 2310$.

My solution:

Note that we could rewrite the given LHS of the inequality as follows:

$35\sqrt{55}+55\sqrt{77}+77\sqrt{35}+35\sqrt{77}+55\sqrt{35}+77\sqrt{55}$

My solution:

Note that we could rewrite the given LHS of the inequality as follows:

$35\sqrt{55}+55\sqrt{77}+77\sqrt{35}+35\sqrt{77}+55\sqrt{35}+77\sqrt{55}$

Prove $35\sqrt{55}+55\sqrt{77}+77\sqrt{35}+35\sqrt{77}+55\sqrt{35}+77\sqrt{55}\gt 2310$.

The first idea that might sprang to your mind whenever you see square root terms and the inequality is you want to squaring both sides of the inequality and you are expected to finally prove a figure is larger than another figure and you are hence done.

The first idea that might sprang to your mind whenever you see square root terms and the inequality is you want to squaring both sides of the inequality and you are expected to finally prove a figure is larger than another figure and you are hence done.

For $a\gt b\gt 0$, prove that $\dfrac{a+b}{2}\gt \dfrac{a-b}{\ln a-\ln b}$.

There sure is many way to prove this problem, but I am going to show you one graphical method that if we recognize some function is always greater than the other in certain interval, then it makes the problem all that easier for us to crack.

There sure is many way to prove this problem, but I am going to show you one graphical method that if we recognize some function is always greater than the other in certain interval, then it makes the problem all that easier for us to crack.

Suppose that there exist two positive real numbers $a$ and $b$ such that $(a-2) (a^2+2a-18)+ab(2a+b+18)-96+2(b-2)(b+3)(b+2)=0$.

Evaluate $a+2b$.

My solution:

Evaluate $a+2b$.

My solution:

Given that $a$ is rational and the equation $ax^2+(a+2)x+a-1=0$ has integer roots.

Find the sum of all possible $a^3$.

It is very important for me to say out loud here that this solution is provided by my math friend, a retired math professor from the U.K.

Find the sum of all possible $a^3$.

It is very important for me to say out loud here that this solution is provided by my math friend, a retired math professor from the U.K.

Given that $a$ is rational and the equation $ax^2+(a+2)x+a-1=0$ has integer roots.

Find the sum of all possible $a^3$.

The solution will require deep and strategic thought but that doesn't mean this problem is impossible to solve.

If you know and are familiar with the manipulation trick that we could play on the rational number, we can see the solution pretty clearly.

Find the sum of all possible $a^3$.

The solution will require deep and strategic thought but that doesn't mean this problem is impossible to solve.

If you know and are familiar with the manipulation trick that we could play on the rational number, we can see the solution pretty clearly.

Quiz 13: Brain Power Enrichment Quiz (II)

The diagram below represents the graph y=x and the coordinate points (2, 2), (4, 4) and (6, 6).

The diagram below represents the graph y=x and the coordinate points (2, 2), (4, 4) and (6, 6).

Labels:
AM-GM inequality,
arc length,
average,
distance,
distance formula,
gradient,
grows faster than,
Jensen's inequlaity,
midpoint,
relation,
strictly increasing,
tangent line,
y=ln(x)

I have heard some very wise saying, it says "What we know isn’t what we need to know. Familiarity, process, and our comfort zones are only holding us back."

Too much familiarity is really bad for creativity. Too much experiences with the common problem may narrow down the choices that you have to deal with the problem at hand, you stick to the old ways of solving problems and you are unable to produce new ideas.

Too much familiarity is really bad for creativity. Too much experiences with the common problem may narrow down the choices that you have to deal with the problem at hand, you stick to the old ways of solving problems and you are unable to produce new ideas.

Solve the following system of equations in real $a,\,b,\,c,\,d$:

$a+b=9$

$ab+c+d=29$

$ad+bc=39$

$cd=18$

Heuristic solution:

$a+b=9$

$ab+c+d=29$

$ad+bc=39$

$cd=18$

Heuristic solution:

Solve the following system of equations in real $a,\,b,\,c,\,d$:

$a+b=9$

$ab+c+d=29$

$ad+bc=39$

$cd=18$

Previously we tried the elimination route and we failed.

$a+b=9$

$ab+c+d=29$

$ad+bc=39$

$cd=18$

Previously we tried the elimination route and we failed.

Solve the following system of equations in real $a,\,b,\,c,\,d$:

$a+b=9$

$ab+c+d=29$

$ad+bc=39$

$cd=18$

$a+b=9$

$ab+c+d=29$

$ad+bc=39$

$cd=18$

Find all positive integers [MATH]n[/MATH] for which [MATH]\sqrt{n+\sqrt{1996}}[/MATH] exceeds [MATH]\sqrt{n-1}[/MATH] by an integer.

My solution:

Let [MATH]\sqrt{n+\sqrt{1996}}-\sqrt{n-1}=k[/MATH], where [MATH]k[/MATH] is a positive integer.

[MATH]\sqrt{n+\sqrt{1996}}=k+\sqrt{n-1}[/MATH]

My solution:

Let [MATH]\sqrt{n+\sqrt{1996}}-\sqrt{n-1}=k[/MATH], where [MATH]k[/MATH] is a positive integer.

[MATH]\sqrt{n+\sqrt{1996}}=k+\sqrt{n-1}[/MATH]

Evaluate $\displaystyle\int^{\dfrac{\pi}{4}}_0 \dfrac{x}{(\sin x+\cos x)\cos x}\ dx$.

Solving for this problem is a breeze if you're someone who is so sensitive about the possibility of the existence of property of symmetry for the integrand function involved.

Solving for this problem is a breeze if you're someone who is so sensitive about the possibility of the existence of property of symmetry for the integrand function involved.

Solve the equation $x+a^3=\sqrt[3]{a-x}$ where a is real parameter.

My solution:

By observation, note that $x=a-a^3$ is a real solution for the equation $x+a^3=\sqrt[3]{a-x}$.

My solution:

By observation, note that $x=a-a^3$ is a real solution for the equation $x+a^3=\sqrt[3]{a-x}$.

Solve the equation $x+a^3=\sqrt[3]{a-x}$ where $a$ is real.

If one wants to do things in haste and quickly solve the above equation for $x$ without thinking much, one would definitely raise both sides of the equation to the third power to get rid of the cube root:

$x+a^3=\sqrt[3]{a-x}$

$(x+a^3)^3=(\sqrt[3]{a-x})^3$

If one wants to do things in haste and quickly solve the above equation for $x$ without thinking much, one would definitely raise both sides of the equation to the third power to get rid of the cube root:

$x+a^3=\sqrt[3]{a-x}$

$(x+a^3)^3=(\sqrt[3]{a-x})^3$

In one of my previous blog posts(optimization-contest-problem), we want to prove that [MATH]\color{yellow}\bbox[5px,blue]{x^4+x^3-x^2-x+1}[/MATH] is always greater than zero for all real $x$, or more specifically, for $x\gt 1$.

In this blog post, we will continue to manipulate the number one to looking for the most efficient and effective solution.

According to Wikipedia (Number One):

One, sometimes referred to as unity, is the integer before two and after zero. One is the first non-zero number in the natural numbers as well as the first odd number in the natural numbers.

According to Wikipedia (Number One):

One, sometimes referred to as unity, is the integer before two and after zero. One is the first non-zero number in the natural numbers as well as the first odd number in the natural numbers.

Subscribe to:
Posts (Atom)