hairyberry

The general formula across line y = x, is \((a, b)\rightarrow (b, a)\), so the result is \((7, 5)\).

But if you want a more general solution that can work for any line, then it is much more difficult, but turning to imaginary numbers + roots of unity is one way to do it.

Set the side lengths of the triangle to be \(x, x, y\)

\(2x+y=5\), so \(2x<5\) and \(x\le2\).

Therefore, because x is a positive integer, x is either 1 or 2.

But we need to be careful because 1, 1, 3 is not a triangle.

So the only possibility is 2, 2, 1.

Apply the heron's formula directly on triangle PQR:

The semiperimeter is \(\frac{36+22+26}{2}=42\).

\([PQR] = \sqrt{42*(6)*(20)*(16)}=\sqrt{80640}=48\sqrt{35}\).

(The [PQR] notation means area).

We know \(\sin(2x)>0, \tan(2x)<0\), therefore x is in quadrant 3. So \(\cos(2x)\)must be negative.

Therefore, \({\cos}^{2}(2x)=1-{\sin}^{2}(2x)=\frac{49}{64}\), \(\cos(2x)=-\frac{7}{8}\).

Becuase cos(2x) is in the third quadrant, then we know x must be in the second quadrant, so cos must be negative

Using our half-angle formula for cos:

\(\cos(x)=-\sqrt{\frac{\frac{1}{8}}{2}}=-\frac{1}{4}\). Therefore cos(x) = -1/4

We know, \(\sin(x)=\cos(\frac{\pi}{2}-x)\).

\(\angle BAD = 90 = \frac{\pi}{2}\), exactly!

\(\because \angle BAC = \angle BAP + \angle PAQ + \angle QAD = \frac{\pi}{2}\)

\(\therefore\angle BAP+\angle QAD=\frac{\pi}{2}-\angle PAQ\)

\(\therefore \sin(\angle PAQ)=\cos(\angle BAP + \angle QAD)=\cos(\angle BAP)\cos(\angle QAD)-\sin(\angle BAP)\sin(\angle QAD)\).

Triangle BAP and QAD are just right triangles, so it's easy to calculate trigonometric values off them.

We get \(\frac{17\sqrt{1189}}{1189}\), as our final answer.

We know:

\(\cos(2x)=1-2 {\sin}^{2}(x)=2{\cos}^{2}(x)-1\).

Using a clever implementation of these formulas,

\(\cos(4x)=2{\cos}^{2}(2x)-1=2{(1-2{\sin}^{2}(x))}^{2}-1\).

We get: \(\cos(4x)=8{\sin}^{4}(x)-8{\sin}^{2}(x)+1\). 

We have a very interesting property, and very useful property when it comes to slope: \(\tan=\frac{O}{A}\), opposite over adjacent, but when put in the coordinate plane, that is \(\tan=\frac{rise}{run}\), and what else is rise over run? Slope! .

Set the angle formed by y=5x and the x axis as \(\theta\), and the angle formed by y=1/6x and the x axis as \(\phi\).

We want to find \(\tan(\phi+\frac{\theta-\phi}{2})=\tan(\frac{\theta+\phi}{2})\).

Use some of our identities:

With a lot of use of tangent identities, we know \(\tan(\theta)=5, \tan(\phi)=1/6\).

Directly applying our tangent angle identity: \(\tan(\theta+\phi)=\frac{5+\frac{1}{6}}{1-\frac{5}{6}}=31\).

Calculate, \(\tan(\theta+\phi)=\frac{\sin(\theta+\phi)}{1+\cos(\theta+\phi)}\).

Draw a triangle and calculate sin and cos.

\(\sin(\theta+\phi)=\frac{31}{\sqrt{962}}\), \(\cos(\theta+\phi)=\frac{1}{\sqrt{962}}\)

Substituting in, we get \(y=(\frac{\sqrt{962}-1}{31})x\) as our line, so \(\bf{m=\frac{\sqrt{962}-1}{31}}\).\(\)

Thanks, I can confirm this is the correct answer, admist this confusion. 

We can verify proyaop's answer through similar triangles \(\triangle BEF \sim\triangle ECD\), so to check we make sure \(\frac{BF}{EF}=\frac{ED}{CD}\), by proyaop's answer, \(BF=3-\frac{3}{4}=\frac{9}{4}\), \(EF=\frac{3}{4}\) , \(ED=\frac{3}{4}\), \(CD=1-\frac{3}{4}=\frac{1}{4}\), so \(\frac{\frac{9}{4}}{\frac{3}{4}}=\frac{\frac{3}{4}}{\frac{1}{4}}=3\).

Asinus did nothing majorly wrong, i think they misread the side lengths as 1 and 2 .  

The probability that at most one of them shows tails = the probability that none of them show tails (all of them are heads) + the probability that one of them shows tails.

For any n, the probability that none of them shows tails is \({(\frac{1}{2})}^{n}\), because heads would needed to be landed on every single time.

The probability that one of these show tails is more difficult. Suppose the first one is tails, and the rest are heads, but we also note that any of these can be the one that lands on heads, so there are n ways to arrage these, so overall the probability is \(n*{(\frac{1}{2})}^{n}\).

Adding these together - \(n*{(\frac{1}{2})}^{n}+{(\frac{1}{2})}^{n}=(n+1){(\frac{1}{2})}^{n}=\frac{5}{16}\), \((n*1)*(\frac{1}{{2}^{n}})=\frac{n+1}{{2}^{n}}=\frac{5}{16}\). We see one such value that satisfies our equality is n = 4. But we know the rate of increase for exponential functions is exponetially increasing/decreasing, so after that the gap between n+1 and 2ⁿ, will only constantly increase, therefore our only positive solution is 4.