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 ABC is an equilateral triangle such that vertices B,C lie on two parallel lines at a distance of 6 units. A lies between the parallel lines at a distance of 4 units from one of them.  Find the side length of the equilateral triangle.

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The side length of a triangle ABC is ≈ 6.11 units

\(y=x^2\\ y=\pm \sqrt{x}\)

The coordinate plane is divided into five planes by the graphs \(y=x^2\) and \(y=\pm\sqrt{x}\).

Solution: \(\mathrm{Range\:of\:}\frac{5x^2+20x+23}{x^2+8x+5}:\quad\begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\le\frac{-\sqrt{421}+16}{11}\quad\mathrm{or}\quad\:f\left(x\right)\ge\frac{\sqrt{421}+16}{11}\:\\ \:\mathrm{Interval\:Notation:}&\:(-\infty\:,\:\frac{-\sqrt{421}+16}{11}]\cup\:[\frac{\sqrt{421}+16}{11},\:\infty\:)\end{bmatrix}\)

Steps: Rewrite \(\frac{5x^2+20x+23}{x^2+8x+5}\)as \(\frac{5x^2+20x+23}{x^2+8x+5}=y\)

\(=\frac{5x^2+20x+23}{x^2+8x+5}=y\)

Multiply both sides by \(x^2+8x+5\).

\(=\frac{5x^2+20x+23}{x^2+8x+5}\left(x^2+8x+5\right)=y\left(x^2+8x+5\right)\)

Simplify \(\frac{5x^2+20x+23}{x^2+8x+5}\left(x^2+8x+5\right):5x^2+20x+23\)

\(=5x^2+20x+23=y\left(x^2+8x+5\right)\)

The range is the set of y for which the discriminant is greater or equal to zero

Discriminant \(5x^2+20x+23=y\left(x^2+8x+5\right):\quad44y^2-128y-60\)

\(44y^2-128y-60\ge\:0\quad:\quad y\le\frac{-\sqrt{421}+16}{11}\quad\) or \(\quad\:y\ge\frac{\sqrt{421}+16}{11}\)

Check if the range interval endpoints are included

\(y=\frac{-\sqrt{421}+16}{11}\quad:\quad\)Included

\(y=\frac{\sqrt{421}+16}{11}\quad:\quad\)Included

Therefore, the range is \(f\left(x\right)\le\frac{-\sqrt{421}+16}{11}\quad\) or \(\quad\:f\left(x\right)\ge\frac{\sqrt{421}+16}{11}\).

Solution #1: If the numbers are called \(x\) and \(y\), then \(y = x - 9\), and so the sum of squares

\(x^2+ y^2 = x^2 + (x-9)^2 = 153\)

\(x^2 + (x^2 - 18x + 81) = 153\)

\(2x^2 - 18x - 72 = 0\)

\(x^2 - 9x - 36 = 0\)

This is factorable. However, if you use the quadratic formula rather than just factoring, the formula shows that either \(x=12\) and \(y=3\), or \(x=-3\) and \(y=-12\). Either way, \(x\cdot y = \boxed{36}\).

Solution #2: \(x-y = 9…(1). x^2+y^2 = 153…(2). (2)-(1)^2\rightarrow2xy = 153–81 =72\text{,} \text{ ie., } xy = \boxed{36}.\)

Both ways, you get the same answer of \(\boxed{36}\).

Solve for a:
a^2 = 3 a + 48

Subtract 3 a + 48 from both sides:
a^2 - 3 a - 48 = 0

Add 48 to both sides:
a^2 - 3 a = 48

Add 9/4 to both sides:
a^2 - 3 a + 9/4 = 201/4

Write the left hand side as a square:
(a - 3/2)^2 = 201/4

Take the square root of both sides:
a - 3/2 = sqrt(201)/2 or a - 3/2 = -sqrt(201)/2

Add 3/2 to both sides:
a = 3/2 + sqrt(201)/2 or a - 3/2 = -sqrt(201)/2

Add 3/2 to both sides:

a = 3/2 + sqrt(201)/2 or a = 3/2 - sqrt(201)/2, OR:

a =~ 8.5887   and    a=~ -5.5887

No problem!

Also, I wanted to notify you that there should be no cheating on Alcumus or Homework as it is not allowed on AoPS.

Thank you for this. If people do need help, they can ask (sorry if im being rude).

Thanks guys! :D

Tysm CPhill!

Just a simple proportion

2 / 100  =   50   /  N          cross-multiply

2N  = 100 * 50

2N =  5000

N  =5000/2   =  2500 salmon

Another way to see this is to  note that 1/50  of the salmon are tagged

So....there are 50 times as  many as we  find  tagged

So...if we  find  50 tagged....then  50 * 50    = 2500  

Assuming that x is a positive ineger

Let x  be  some "large positive integer "

The ordering of the  numbers  is

2, 3 , 5 , x , 2x

And the median is  5.....just  as   cryptoaops   found  !!!

 Note  that not just  ANY integer will produce this....for example...if  x  = 1   we have

1 2 2 3 5    and the median is  2  

And if x  = 2  we  have    2  2  3  4  5  and the median    = 3

If x =   3  we have   2 3 3  5  6       median  = 3

If x =  4  we have   2 3 4 5  8       median = 4

In fact  x  must be an integer greater than  4 to have a median of 5

10=(a³+b³)

    =(a+b)(a²-ab+b²)                  Factor differnce of cubes

    =(a+b)(a²+2ab+b²-3ab)

    =(a+b)((a+b)²-3ab)

    =(2)(2²-3ab)

so

5=4-3ab

1=-3ab

ab=-1/3

\(2, 3, 5, x, 2x\) creates the largest median no matter what. \(x\) can be any integer you want it to be, but the median will always be \(5\)

If we inscribe  the hexagon in a circle, let  the radius  = R

By AAS  we  have 8 congruent  30-60-90  right triangles

The side opposite the 60° angle = R/2

And the side opposite  the   30°  angle =  R / ( 2sqrt (3) )

So.....the area  of  these  8 right triangles  is

8 *(1/2) (R/2) (R  /(2sqrt (3))  =   R^2/sqrt (3)

And  the  side of the  hexagon  = R....so its area =  3sqrt (3)R^2 /2

So   H  /  S    =        3sqrt(3) R^2 / 2               3 sqrt (3) sqrt (3)             9

                              _____________  =          ______________   =     ____

                               R^2 / sqrt (3)                             2                            2

Exactly what   cryptoaops found  !!!!

Good job, cryptoaops  !!!!

The answer is actually 7.9, but thnx!

First of all, the image is below not above 

The hexagon is composed of 6 isoceles triangles. In the sketch above, you just have to draw their apothem and then you get 6x2 right-angled triangles. So you have the same little right-angled triangles composing the whole hexagone. Each isoceles triangle has 6 little right-angled triangles and the total is \(\LaTeX\): \(6\text{isoceles}\triangle\times6\text{little right-angled triangle}=36\). You count \(\LaTeX\): \(4\text{little right-angled}\triangle\times2\text{shaded area}=8\text{shaded little right-angled triangles}\). So the ratio is: \(\frac{36}{8}=\frac{9\times4}{2\times4}=\boxed{\frac92}\). And \(\dfrac92=9:2\).

Since it is a ratio question you can use a convenient ellipse.

Let X be the centre and M be the midpoint of PR

triangle QXF2    and  triangle QMP   are similar tiranges and they are both 30, 60 90 degree triangles.

So their sides are in the ration 1:2:sqrt3

We can let the focal length (c) to be 1 and we can Q be (0,1).  Then a = 2.

From here, with routine calculations, you can work out that PQ/F_1 F_2 = 13/8.

We see that z=y+4

using Pythagorean theorem, we find that (x+y)^2+(y+z)^2=(x+z)^2, 

plugging in the equation for 2 variables, we get 

(x+y)^2+(2y+4)^2=(x+y+4)^2

if we simplify the above equation of two variables, then you should get

y^2+2y=2x

now use some trial and error on y and see if x and z will satisfy, hint: there is only one solution

A(1)  =  p(1)   + q^1  =   p +  q  =  4  ⇒      q =   4 - p      (1)

4A(2)  = 4 [ p(2)  + q^2 ] =  8p + 4q^2  =  37         (2)

Sub (1)  into (2)  for  q 

8p + 4 (4 - p)^2    = 37

8p + 4 (p^2   - 8p +16)  =  37

8p + 4p^2  - 32p + 64   = 37

4p^2  - 24p + 27  =  0        factor

(2p - 9)(2p - 3)  =  0

Set each factor to  0  and  solve  for  p   and we  get that  p =  9/2    or  p  =3/2

When  p =  9/2, q  =  4 - (9/2)  = -1/2   and     p  - q =  9/2 -  - 1/2  =  10/2  = 5

When p=  3/2 , q = 4 - (3/2)  = 5/2   and p - q  =  3/2 - 5/2  =  -2/2  =   - 1

So max value  of    p - q   =    5

(3,7)  (6, 2)   ( 2, k)

We need to find the slope  between   (3,7)   and (6,2)

(7 - 2)   / ( 3  - 6)  =  -5/3

Now  we need to find  two  lines with  reciprocal slopes  ( perpendicular to AB)  that pass through (3,7)  and (6,2)

So...one line has the equation

y = (3/5) ( x - 3)  + 7

y = (3/5)x - 9/5 + 35/5

y = (3/5)x + 26/5

So one value of  k  is

k = (3/5)(2) + 26/5

k = 6/5 + 26/5

k = 32/5

And  the  other line has the equation

y = (3/5)( x - 6)  + 2

y = (3/5)x - 18/5 + 10/5

y = (3/5)x -8/5

So  the other value of  k is

k  = (3/5)(2)  - 8/5

k = 6/5 - 8/5

k  = -2/5