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Let triangle ABC have side lengths AB=13, AC=14, and BC=15.

There are two circles located inside angle BAC which are tangent to rays AB, AC, and segment BC.

Compute the distance between the centers of these two circles.

\(\begin{array}{|rcll|} \hline s &=& \dfrac{a+b+c}{2} \\ s &=& \dfrac{13+15+14}{2} \\ \mathbf{s} &=& \mathbf{ 21 } \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline r &=& \sqrt{\dfrac{(s-a)(s-b)(s-c)}{s}} \\ &=& \sqrt{\dfrac{(21-13)(21-15)(21-14)}{21}} \\ &=& \sqrt{\dfrac{(8)(6)(7)}{21}} \\ &=& \sqrt{\dfrac{336}{21}} \\ &=& \sqrt{16} \\ \mathbf{r} &=& \mathbf{ 4 } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline r_b &=& \sqrt{\dfrac{s(s-a)(s-c)}{s-b}} \\ &=& \sqrt{\dfrac{21(21-13)(21-14)}{21-15}} \\ &=& \sqrt{\dfrac{21(8)(7)}{6}} \\ &=& \sqrt{ 196 } \\ \mathbf{r_b} &=& \mathbf{ 14 } \\ \hline \end{array}\)

\(\mathbf{AU=\ ?}\)

\(\begin{array}{|rcll|} \hline AU &=& AV \\ AU+AV &=& (a+BU)+(c+CV) \\ &=& a+c+BU+CV \quad | \quad BU = BU',\ CV=CV' \\ &=& a+c+BU'+CV' \quad | \quad BU'+CV' = b \\ &=& a+c+ b \\ AU+ AV&=& 2s \quad | \quad AV = AU \\ 2AU &=& 2s \\ \mathbf{ AU } &=& \mathbf{ s } \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline x &=& \sqrt{\Big(s-(s-b)\Big)^2 + (r_b-r)^2 } \\ &=& \sqrt{b^2 + (14-4)^2 } \\ &=& \sqrt{15^2 + 10^2 } \\ &=& \sqrt{(3*5)^2 + (2*5)^2 } \\ &=& \sqrt{5^2*(3^2+2^2) } \\ &=& \sqrt{5^2*13 } \\ \mathbf{x } &=& \mathbf{ 5\sqrt{ 13 } } \\ x &\approx& 18 \\ \hline \end{array} \)

The distance between the centers of these two circles is \(\mathbf{\approx 18}\)

If -5\leq a \leq -1$ and $1 \leq b \leq 3$, what is the least possible value of $\displaystyle\left(\frac{1}{a}+\frac{1}{b}\right)\left(\frac{1}{b}-\frac{1}{a}\right) $? Express your answer as a common fraction.

Tim is doing a cakewalk with 11 steps. He takes his first step on step 1 and takes a total of 139 steps,
walking in a circle (so that after the 11th step he reaches the first step again).
Which step of the cakewalk does he end on?

I assume:

\(\begin{array}{|rcll|} \hline 139 &=& \underbrace{12\times 11}_{\text{position }1} + 7 \\ \text{Tim end on position } 1 + 7 = 8 \\ \hline \end{array}\)

Let $S$ be the set of numbers of the form \[n(n + 1)(n + 2)(n + 3)(n + 4),\] where $n$ is any positive integer.

The first few terms of $S$ are \begin{align*} 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 &= 120, \\ 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 &= 720, \\ 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 &= 2520, \end{align*} and so on.

What is the GCD of the elements of $S$?

I assume:

The GCD of the elements of \(S\) is 120.

\(\begin{align*} 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 &= 120 = 2^3\times3\times5\times7^0\times 11^0\times 13^0\times ... \times p^0 \\ \end{align*}\)

 All other values contain a multiple of 4 and another even number, so at least a \(2^3\)
 All other values contain a multiple of 3, so at least a \(3^1\)
 All other values contain a multiple of 5, so at least a \(5^1\), while there are 5 concecutive numbers.
The gcd is the product of all prime numbers with the smallest exponent. The exponent zero gives the factor 1.

So this is the value of the first therm.

 Continued Fraction Problem

Hallo mathismyhobby

Man erkennt das am gleichen Parameter und an deren Vorzeichen.

Z.B.  -6y und -6y kürzen sich zu Null bei Subtraktion.

        -6y und +6y kürzen sich zu Null bei Addition.

Ich scheitere an dieser Aufgabe beim "horizontal addieren"

In diesem Fall empfielt sich die Subtraktion:

\(\begin{array}{|rcll|} \hline \mathbf{5x-6y}&=&\mathbf{3} \\ \mathbf{7x-6y}&=&\mathbf{6} \\\\ (5x-6y)-(7x-6y)&=&(3)-(6) \\ 5x-6y-7x+6y &=&-3 \\ 5x-7x-6y+6y &=&-3 \\ -2x &=&-3 \\ 2x &=& 3 \\ \mathbf{x} &=& \mathbf{\dfrac{3}{2}} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \mathbf{5x-6y}&=&\mathbf{3} \\\\ 6y &=& 5x-3 \quad | \quad \mathbf{x=\dfrac{3}{2}} \\ 6y &=& 5\left(\dfrac{3}{2}\right)-3 \\ 6y &=& \dfrac{15}{2} -3\left(\dfrac{2}{2}\right) \\ 6y &=& \dfrac{15}{2} - \dfrac{6}{2} \\ 6y &=& \dfrac{15-6}{2} \\ 6y &=& \dfrac{9}{2} \quad | \quad : 6 \\ y &=& \dfrac{9}{12} \\\\ \mathbf{y} &=& \mathbf{\dfrac{3}{4}} \\ \hline \end{array}\)

Find \(2^{-1} \pmod{185}\), as a residue modulo 185. (Give an answer between 0 and 184, inclusive.)

According to Euler's theorem, if a is coprime to m, that is, gcd(a, m) = 1, then
  \( {\displaystyle a^{\phi (m)}\equiv 1{\pmod {m}},}\)
where   \({\displaystyle \phi } \) is Euler's totient function.  
A modular multiplicative inverse can be found directly:
   \({\displaystyle a^{\phi (m)-1}\equiv a^{-1}{\pmod {m}}.}\)

2 must be coprime to 185 or \(gcd(2, 185) = 1~ \checkmark\)

\(\begin{array}{|rcll|} \hline && 2^{-1}{\pmod {185}} \quad & | \quad 185 = 5\cdot 37 \\ &\equiv& 2^{\phi (185)-1}{\pmod {185}} \quad &| \quad \phi (185) = 185\left(1-\dfrac{1}{5} \right)\left(1-\dfrac{1}{37} \right)= 144 \\ &\equiv& 2^{144-1}{\pmod {185}} \\ &\equiv& 2^{143}{\pmod {185}} \quad & | \quad 2^{11} = \mathbf{13} \pmod{185} \\ &\equiv& 2^{11*13}{\pmod {185}} \\ &\equiv& \left(2^{11} \right)^{13} {\pmod {185}} \\ &\equiv& \mathbf{13}^{13} {\pmod {185}} \quad & | \quad 13^{5} = \mathbf{-2} \pmod{185} \\ &\equiv& 13^{5*2+3}{\pmod {185}} \\ &\equiv& \left(13^{5} \right)^{2}*13^3 {\pmod {185}} \\ &\equiv& \left(\mathbf{-2}\right)^{2}*13^3 {\pmod {185}} \\ &\equiv& 4*13^3 {\pmod {185}} \\ &\equiv& 8788 {\pmod {185}} \\ &\equiv& \mathbf{93} {\pmod {185}} \\ \hline \end{array}\)

The modular multiplicative inverse of 2 is 93

\(Proof: 2 * 93 = 186 \equiv 1 \pmod{185} ~ \checkmark \)

 Question from Homework

    Do you see the Pascal's triangle ?

The number of paths form A to B that pass through C is \(\mathbf{10*4 = 40}\).

There's some number of triangles satisfying
What is the sum of all the possible values of BC? If there are no possible values, answer with 0.

\(\mathbf{\text{cos-rule}}:\)

\(\begin{array}{|rcll|} \hline a^2 &=& b^2+c^2-2bc\cos(A) \\ \mathbf{\cos(A)} &=& \mathbf{\dfrac{b^2+c^2-a^2}{2bc}} \\ \hline \end{array} \)

\(\mathbf{\text{projection-rule}}:\)

\(\begin{array}{|rcll|} \hline b &=& c\cos(A)+a\cos(C) \quad | \quad \mathbf{\cos(A)=\dfrac{b^2+c^2-a^2}{2bc}} \\ b &=& c\left(\dfrac{b^2+c^2-a^2}{2bc} \right)+a\cos(C) \\ b &=& \dfrac{b^2+c^2-a^2}{2b } +a\cos(C)\quad | \quad *2b \\ 2b^2 &=& b^2+c^2-a^2 +2ab\cos(C) \\ b^2 &=& c^2-a^2 +2ab\cos(C) \\ \mathbf{a^2 -2ab\cos(C) +b^2-c^2} &=& \mathbf{0} \\\\ a &=& \dfrac{2b\cos(C)\pm \sqrt{4b^2\cos^2(C)-4*(b^2-c^2)} }{2} \\ a &=& b\cos(C) \pm \dfrac{\sqrt{4b^2\cos^2(C)-4*(b^2-c^2)} }{2} \\ a_1+a_2 &=& 2b\cos(C) \quad | \quad b=10,\quad C = \dfrac{\pi}{6} \\ &=& 2*10\cos(\dfrac{\pi}{6}) \quad | \quad \cos(\dfrac{\pi}{6}) = \dfrac{\sqrt{3}}{2} \\ &=& 2*10*\dfrac{\sqrt{3}}{2} \\ &=& 10 \sqrt{3} \\ \mathbf{a_1+a_2} &=& \mathbf{17.3205080757} \\ \hline \end{array}\)

The sum of all the possible values of BC is 17.3205080757