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Hi CuriousDude and Max,

I really like these. Thanks.

I'll add a link to them in our Latex stickt post.   

I checked and there are no mistypes.  I have changed it to latex so you can look at it easier.

Give your answer in "standard form"  ax+by+c>0  or   ax+by+c>=0 and a b c are integers with no common factor other than 1. I’m sorry I forgot about the format. Help! How do I convert it?

Equation of the red line is     y = -x +1

the shaded area PLUS the included red line :    y <= -x+1

Formatted:    y+x -1 <= 0      or   -y-x+1>=0

\(\begin{cases} 2x + 5y = -8\\ 6x + 15y = 16 + a \end{cases}\\ \begin{cases} 2x + 5y = -8\\ 2x + 5y = \dfrac{16 + a}3 \end{cases}\\ \dfrac{16+a}{3} = -8\\ 16 + a = -24\\ a = -40\)

\(P(x) = x^3 + bx^2 + cx + d \\ (2^{1/3}+1)^3 + b(2^{1/3}+1)^2 + c(2^{1/3}+1) +d = 0\\ 3 + 3\left(2^{2/3}+2^{1/3}\right) + b(2^{2/3})+2b(2^{1/3})+b+c(2^{1/3})+c+d=0\\ (3+b)(2^{2/3}) + (3+2b+c)(2^{1/3}) + (3+b+c+d) = 0\\ 3 + b = 0\\ b = -3\\ 3 + 2b + c = 0\\ 3 - 6 + c = 0\\ c = 3\\ 3 + b + c + d = 0\\ 3 - 3 + 3 + d = 0\\ d = -3\\ \therefore \text{The required monic polynomial is } x^3 -3x^2+3x-3=0 \)

Factors of 2^95 = {1,2,4,8,16,...,2^94,2^95}.

The first power of 2 that is greater than 1,000,000 is 2^20, which is 1.048,576.

Number of factors of 2^95 which are greater than 1,000,000 = 95 - 20 + 1 = 76

A triangle must have positive area. The only case it doesn't, is when the three points are collinear(on the same straight line).

And, your math teacher isn't evil. There are only 10 lattice points for us to choose. 

We only need to consider how many combinations of those 3 points are collinear, subtract that from the total number of possible combinations of choosing 3 points out of 10 points, and we get our answer.

Firstly, number of ways we can choose 3 points from 10 points = \(^{10}C_3 = 120\).

Consider any straight line that passes through 3 or more points of the pattern.

There are 2 vertical lines, 2 horizontal lines, and 2 oblique lines that passes through 3 or more points of the pattern.

One of each type of lines passes through 4 points of the pattern, and another of each type of lines passes through 3 points of the pattern.

Let's call those lines which passes through 4 points "Type A" lines, and those which doesn't "Type B" lines.

Number of ways that all 3 points lies on a "Type A" line = \(^4C_3 = 4\)

There are 3 "Type A" lines: \(3\left(^4C_3\right) = 12\)

Number of ways that all 3 points lies on a "Type B" line = \(^3C_3 = 1\)

There are 3 "Type B" lines: \(3\left(^3C_3\right) = 3\)

Now, subtract these cases from the total number of cases.

Number of ways that 3 points in the pattern are chosen such that they form a triangle with positive area

= 120 - 12 - 3

= 105

In case you don't know these, you can list the possible ways and count them one by one. There are only 105 cases so it should be fast enough.

\(\begin{array}{rcll} 2x^2 &=& 4x + 9\\ 2x^2 - 4x - 9 &=& 0\\ x &=&\dfrac{4\pm\sqrt{16+4\cdot2\cdot9}}{4}\\ x &=& \dfrac{4\pm2\sqrt{22}}4\\ x &=& \dfrac{2\pm\sqrt{22}}2\\ a + b + c &=&2 + 22 + 2 \\ a + b + c &=& 26 \end{array}\)

\(\begin{array}{rcll} x(x-3) &=& 1\\ x^2 - 3x &=& 1\\ x^2 - 3x - 1 &=& 0\\ x &=& \dfrac{3\pm \sqrt{3^2 + 4}}{2}\\ x &=& \dfrac{3\pm\sqrt{13}}{2}\\ abc &=&3\cdot 13 \cdot 2 \\ abc &=& 78 \end{array}\)

The exact same question has been asked here before.

Denote the point of tangency as T.

To attain minimum, AO and OB must be as short as they can be. As the unit circle is symmetric, AO must be equal to OB when minimum of AB is attained.

\(\text{AO} = \text{OB} \wedge\angle \text{AOB} = 90^{\circ}\).

When minimum of AB is attained, \(\triangle \text{AOB}\) is a 90-45-45 triangle.

OT = 1 unit

As \(\triangle \text{AOB}\) is an isosceles triangle, the radius joining O to the point of tangency is a perpendicular bisector and it divides \(\triangle \text{AOB}\) into 2 equal halves, which are two 90-45-45 triangles.(If you can't understand this, go to the kitchen, grab a square piece of bread, cut it along the main diagonal, and then cut one of the two parts into half.)

As \(\triangle \text{AOT}\) is a 90-45-45 triangle with angle OTA = 90 degrees, TA = TO = 1 unit.

As the shape is symmetric along the line OT, TB = TA = 1 unit.

So when minimum is attained, AB = TA + TB = 1 + 1 = 2 units

\(\begin{array}{rcll} 5x - 17 &=& -x+7&|\quad\text{Add x to both sides}\\ 6x-17&=&7&|\quad\text{Add 17 to both sides}\\ 6x&=&24&|\quad\text{Divide both sides by 6}\\ x&=&4&|\quad\color{red}\text{FINISH!} \end{array}\)

\(\text{AD}^2 + \text{DC}^2 = \text{AC}^2\\ \text{AD}^2 = 13^2 - 5^2 = 144\\ \text{AD} = 12\)

\(\text{BD}^2 + \text{DA}^2 = \text{AB}^2\\ \text{BD}^2 = 15^2 - 12^2 = 81\\ \text{BD} = 9\\ \text{BC} + \text{CD} = 9\\ \text{BC} = 9 - 5 = 4\)

Area = \(\dfrac{4\times 12}{2} \) = \(24 \;\text{sq. units}\)

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Beat me in this :P

Script: 

\color{red}{\LaTeX}^{\color{BurntOrange}{\LaTeX}^{\color{Dandelion}{\LaTeX}^{\color{ForestGreen}{\LaTeX}}\color{CornflowerBlue}{\LaTeX}}\color{RoyalBlue}{\LaTeX}}_{\color{BurntOrange}{\LaTeX}_{\color{Dandelion}{\LaTeX}_{\color{ForestGreen}{\LaTeX}}\color{CornflowerBlue}{\LaTeX}}\color{RoyalBlue}{\LaTeX}}\color{purple}{\LaTeX}

In the coordinate plane, let A be a point on the x-axis, and let B be a point on the y-axis, so that AB is tangent to the unit circle. Find the minimum value of AB.

The point of contact let P be, the origin of the coordinate plane O.
Triangle AOB is similar to triangle \(PX_PO\).

\(A_{PX_PO}=\frac{x}{2}\cdot y\)

\(y=\sqrt{1-x^2}\)

\(A=\frac{x}{2}\sqrt{1-x^2}\\ A=\frac{1}{2}\sqrt{x^2-x^4}\)

\(A=\frac{1}{2} (x^2-x^4)^{\frac{1}{2}}\)

\(A'=\frac{1}{2}(2x-4x^3)\cdot \frac{1}{2}(x^2-x^4)^{-\frac{1}{2}}=0\\ A'=\frac{2x-4x^3}{4(x^2-x^4)^{\frac{1}{2}}}=0\\ A'=\frac{x-2x^3}{2(x^2-x^4)^{\frac{1}{2}}}=0\\ \color{blue}x=\frac{1}{\sqrt{2}}\)

\(x=0,7071067811865476\)

\(A=\frac{1}{2}\sqrt{x^2-x^4}\)

\(A_{X_PO}=0.5\)

\(A_{AOB}=​4\cdot A_{X_PO}=2 ​\)

The smallest area of the triangle AOB is equal to 2.

  !

Now you try

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Which is always why you read the problem right! It's ok, people make mistakes

I'm pretty sure that you mistyped that; the answer is unknown.