hairyberry

Nice!

I loved how you included the units in your calculation. That helps the understanding of the problem a lot. 

A Nice trick, that some people don't know about, is that you can just directly multiply the units. 

For in example \(3 hour * \frac{2 miles}{hour} = 6 miles\), the hours cancel out. Just remember to multiply by the correct unit fraction, because \(3 hour * \frac{hour}{2miles}=\frac{3{hour}^2}{2miles}\), and while equivalent, not our desired unit. This can done multiple times to get any desired unit, which can be very useful in rate problems.

Happy Solving!

We see trigonmetry might be a possible solution to the problem, but here is a non-trigonmetry solution.

Instead, we focus on the fact that this triangle is an isosceles triangle, so we can draw a perpendicular to solve this problem.

Consider the diagram below.

We can now use a system of equations to solve this problem. \(\sqrt{{x}^{2}+{y}^{2}}=7, \frac{2xy}{2}=14\)

Simplifying, we get \({x}^{2}+{y}^{2}=49, xy=14\). This is not an easy system to solve, but there is a small trick I often use for a problem like this. We recognize the structure of \(x+y=\sqrt{{x}^{2}+2xy+{y}^{2}}=\sqrt{49+14*2}=\sqrt{77}\). Now we can use a little trick with the Vieta's formulas. We recognize that x and y are the two roots of the equation \({a}^{2}-\sqrt{77}a+14\). (because the sum of the roots is sqrt(77) and the product of the roots is 14). We recoginze here, that x and y are interchangable, because we can assign any of the roots to x or y.

But we need to read the problem. It is not asking the roots, but the product of the perimeters. Because we know x and y are the two roots, or the two solutions to the base, we calculate \((14+2x)(14+2y)=196+28(x+y)+4xy=196+28\sqrt{77}+56=252+28\sqrt{77}\). So our answer is 252+28sqrt(77).

Here is a solution that doesn't require any complex geometry. We see that we know all 3 side lengths of triangle BID. We know a formula that can tell us the area directly knowing 3 sides of the triangle, Heron's formula.

First we calculte the semi-perimeter \(\frac{3+4+6}{2}=\frac{13}{2}\). 

Then we apply Heron's formula. \(Area = \sqrt{\frac{13}{2}(\frac{13}{2}-3)(\frac{13}{2}-4)(\frac{13}{2}-6)}\). Simplifying, we get \(Area = \sqrt{\frac{455}{16}}=\frac{\sqrt{455}}{4}\) 

So, the Area is sqrt(455)/4 which is around 5.3327.

Thanks!!!

Hello,

To convert \(0.038\overline{146}\) into a fraction, we can use a relatively simple operation, which can be used for solving a variety of problems. We set \(x = 0.038\overline{146}\), but we also see that mutitplying x by 1000 produces a similar repeating decimal, \(1000x = 38.\overline{146}\). We see that these repeating decimals, are both some relatively simple number + the repeating decimal 0.000146146.... To cancel out, we subtract x from 1000x, yielding \(1000x-x=38.\overline{146} - 0.038\overline{146} \), so \(999x=38.108\), and after simplification, x=9527/249750

We recognize that these expressions both contain the terms \({a}^{3}\) and \(a{b}^{2}\), so we can set these as variables, let \(x = {a}^{3}\) and \(y = a{b}^{2}\).

We now turn this into a linear system of equations, \(x+3y=679\) and \(3x-y=615\)

We solve this system of equations, getting \(x, y=\frac{1262}{5}, \frac{711}{5}\). \(\) 

The Solution eventually is very ugly, but it is \(a-b=\sqrt[3]{\frac{1262}{5}}-\sqrt{\frac{\frac{711}{5}}{\sqrt[3]{\frac{1262}{5}}}}\) which is approximately 1.57617067733.

We can convert these two seperately. 

0.001001001... when expressed in Base 2 is \(0*{2}^{-1}+0*{2}^{-2}+1*{2}^{-3}+0*{2}^{-4}+0*{2}^{-5}+1*{2}^{-6}...\)

The terms with the 0's are all 0, so this simplifies into \({2}^{-3}+{2}^{-6}+{2}^{-9}+{2}^{-12}...\)

Using the Infinite geometric series formula, (first term)/(1 - common ratio), we get \(\frac{{2}^{-3}}{1-{2}^{-3}}\)which simplifies to \(\frac{1}{7}\).

We do the second one similarly, simplifying 0.6666666666666 base 8 into \(6*{8}^{-1}+6*{8}^{-2}+6*{8}^{-3}+6*{8}^{-4}\), using the infinite geometric series formula, we get, \(6*\frac{{8}^{-1}}{1-{8}^{-1}}\)which simplifies to \(\frac{6}{7}\). Adding, 1/7 and 6/7 we get 1.

Label the tangent point between Semicircle AB and the red circle as D, the tangent point between arc BC and the red circle as E, and the tangent point between the red circle and semicircle AC as F.  Label the center of semicircle AB to C₁ the center of semicircle of BC to C₂ and the radius of the red circle to C₃.

Additionally, set the radius of the red circle is x.

We know BF is 2, and so BC₃ is 2-x. 

Similarly, C₁C₃ is 1+x.

Construct the right triangle, C₁C₃B.

Using the pythagorean theorem, we can set an equation:

\({1}^{2}+{(2-x)}^{2}={(1+x)}^{2}\). In the equation, the x's conveniently cancel out, and we get radius = 2/3.

\(cos( \theta)+isin(\theta) = {e}^{i\theta}\) by Euler's formula, or representing the point on the unit circle with terminal side with degree \(\theta\). 

Similarly,

 \(sin(\theta)+icos(\theta) = cos(\pi/2 - \theta) + isin(\pi/2 - \theta) = {e}^{i(\pi/2 - \theta)}\), and this represents the angle with terminal side with degree \(\pi/2 - \theta\).

If a complex number is a root of a polynomial, we also know that the conjugate is a root of a polynomial.

Therefore, we know that, \({e}^{-i\theta}\)and \({e}^{i(\theta-\pi/2)} = {e}^{i(3\pi/2 + \theta)}\)are also roots.

We see that these roots form a trapezoid. (stuff up there not completely necessary. it was my first instinct).

The formula for the area of a trapezoid, is (top + bottom)*height/2.

Finding these values, and plugging in, \(Area = (2sin(\theta)+2cos(\theta))*(cos(\theta) - sin(\theta))/2 = {cos}^{2}(\theta)-{sin}^{2}(\theta)=cos(2\theta)\)

First we know that P(0) is equal to the last term, or the constant term, which is also equal to the product of the roots by Vietas,

Multiplying, \({e}^{i\theta}*{e}^{-i\theta}*{e}^{i(\pi/2-\theta)}*{e}^{i(\theta-\pi/2)} = {e}^{0} = 1\)

So P(0) is equal to 1

The Area is half of P(0) which is 1/2 so \(cos(2\theta)=1/2\)

\(2\theta = \pi/3\)

\(\theta=\pi/6\)

The sum of the roots is, \(2cos(\pi/6)+2sin(\pi/6)=\sqrt{3}+1\)

All real numbers is correct

|x-2| represents the distance between 2, and |x-7| represents the distance between 7.

The sum of the distance between a number and 2 and a number and 7 is a minimum of 5. 

So x is all real numbers.