heureka

In triangle PQR, PQ=13, QR=14, and PR=15.
Let M be the midpoint of QR.
Find PM.

cos Rule:
\(\begin{array}{|rcll|} \hline 15^2&=&13^2+14^2-2*13*14*\cos(Q) \\ 2*13*14*\cos(Q) &=& 13^2+14^2-15^2 \\ 2*13*14*\cos(Q) &=& 140 \quad | \quad : 2 \\ 13*14*\cos(Q) &=& 70 \qquad (1) \\ \hline \end{array}\)

cos Rule:
\(\begin{array}{|rcll|} \hline PM^2 &=& 13^2+\left(\dfrac{14}{2}\right)^2-2*13*\dfrac{14}{2}*\cos(Q) \\\\ PM^2 &=& 13^2+\left(\dfrac{14}{2}\right)^2-13*14*\cos(Q) \\\\ 13*14*\cos(Q) &=& 13^2+\left(\dfrac{14}{2}\right)^2-PM^2 \\\\ 13*14*\cos(Q) &=& 13^2+7^2-PM^2 \\\\ 13*14*\cos(Q) &=& 218-PM^2\qquad (2) \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (1)=(2):& 70 &=& 218-PM^2 \\ & PM^2&=& 218 - 70 \\ & PM^2&=& 148 \\ & PM^2&=& 4*37 \\ & \mathbf{PM}^2 &=&\mathbf{ 2\sqrt{37} } \\ \hline \end{array}\)

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.

Given positive integers x and y such that \(x \ne y\) and

\( \dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{18}\),
what is the smallest possible value for \(x+y\)?

\(\begin{array}{|rcll|} \hline \dfrac{1}{x} + \dfrac{1}{y} &=& \dfrac{1}{18} \\\\ \dfrac{x+y}{xy} &=& \dfrac{1}{18} \\\\ \mathbf{xy} &=& \mathbf{18*(x+y)} \\ \hline \end{array}\)

\(\large{AM\ge GM}\)

\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{x+y}{2}} &\ge& \mathbf{\sqrt{xy}} \\ x+y &\ge& 2\sqrt{xy} \quad &| \quad \text{square both sides} \\ (x+y)^2 &\ge& 4xy \quad | \quad xy = 18*(x+y) \\ (x+y)^2 &\ge& 4*18*(x+y) \\ x+y &\ge& 4*18 \\ \mathbf{x+y } &\ge& \mathbf{72} \\ \hline \end{array}\)

For positive real numbers and the equation
\(\arctan x + \arccos \frac{y}{\sqrt{1 + y^2}} = \arcsin \frac{3}{\sqrt{10}}\)
reduces to an equation of the form \(xy + ax + by + c = 0\).

My attempt:

\(\small{ \begin{array}{|rcll|} \hline \arctan x + \arccos \frac{y}{\sqrt{1 + y^2}} &=& \arcsin \frac{3}{\sqrt{10}} \quad | \quad \tan()~ \text{both sides} \\\\ \tan \Big(\arctan x + \arccos \frac{y}{\sqrt{1 + y^2}} \Big) &=& \tan \Big( \arcsin \frac{3}{\sqrt{10}} \Big) \\\\ \dfrac{ \sin \Big(\arctan x + \arccos \frac{y}{\sqrt{1 + y^2}} \Big) } { \cos \Big(\arctan x + \arccos \frac{y}{\sqrt{1 + y^2}} \Big) } &=& \dfrac{ \sin \Big( \arcsin \frac{3}{\sqrt{10}} \Big) } { \cos \Big( \arcsin \frac{3}{\sqrt{10}} \Big) } \\\\ \hline \dfrac{ \sin (\arctan x) \cos( \arccos \frac{y}{\sqrt{1 + y^2}} ) + \cos (\arctan x) \sin( \arccos \frac{y}{\sqrt{1 + y^2}} )} { \cos (\arctan x) \cos( \arccos \frac{y}{\sqrt{1 + y^2}} ) - \sin (\arctan x) \sin( \arccos \frac{y}{\sqrt{1 + y^2}} )} &=& \dfrac{ \sin \Big( \arcsin \frac{3}{\sqrt{10}} \Big) } { \cos \Big( \arcsin \frac{3}{\sqrt{10}} \Big) } \\\\ \hline \dfrac{ \sin (\arctan x) \frac{y}{\sqrt{1 + y^2}} + \cos (\arctan x) \sin( \arccos \frac{y}{\sqrt{1 + y^2}} )} { \cos (\arctan x) \frac{y}{\sqrt{1 + y^2}} - \sin (\arctan x) \sin( \arccos \frac{y}{\sqrt{1 + y^2}} )} &=& \dfrac{ \frac{3}{\sqrt{10}} } { \cos \Big( \arcsin \frac{3}{\sqrt{10}} \Big) } \\\\ \boxed{\sin (\arctan x) = \frac{x}{\sqrt{1 + x^2}} \\ \cos (\arctan x) = \frac{1}{\sqrt{1 + x^2}} } \\ \dfrac{ \frac{x}{\sqrt{1 + x^2}} \frac{y}{\sqrt{1 + y^2}} + \frac{1}{\sqrt{1 + x^2}} \sin( \arccos \frac{y}{\sqrt{1 + y^2}} )} { \frac{1}{\sqrt{1 + x^2}} \frac{y}{\sqrt{1 + y^2}} - \frac{x}{\sqrt{1 + x^2}} \sin( \arccos \frac{y}{\sqrt{1 + y^2}} )} &=& \dfrac{ \frac{3}{\sqrt{10}} } { \cos \Big( \arcsin \frac{3}{\sqrt{10}} \Big) } \\\\ \boxed{\sin( \arccos z ) = \sqrt{1 - z^2} \\ \cos( \arcsin z ) = \sqrt{1 - z^2} } \\ \dfrac{ \frac{x}{\sqrt{1 + x^2}} \frac{y}{\sqrt{1 + y^2}} + \frac{1}{\sqrt{1 + x^2}} \sqrt{1 - \frac{y^2}{1 + y^2} } } { \frac{1}{\sqrt{1 + x^2}} \frac{y}{\sqrt{1 + y^2}} - \frac{x}{\sqrt{1 + x^2}} \sqrt{1 - \frac{y^2}{1 + y^2} } } &=& \dfrac{ \frac{3}{\sqrt{10}} } { \sqrt{1- \frac{9}{10} } } \\\\ \hline \dfrac{ \frac{x}{\sqrt{1 + x^2}} \frac{y}{\sqrt{1 + y^2}} + \frac{1}{\sqrt{1 + x^2}} \frac{1}{\sqrt{1 + y^2}} } { \frac{1}{\sqrt{1 + x^2}} \frac{y}{\sqrt{1 + y^2}} - \frac{x}{\sqrt{1 + x^2}} \frac{1}{\sqrt{1 + y^2}} } &=& \dfrac{ \frac{3}{\sqrt{10}} } { \frac{1}{\sqrt{10}} } \\\\ \hline \dfrac{xy+1}{y-x} &=& 3 \\\\ xy+1 &=& 3(y-x) \\\\ xy-3(y-x)+1 &=& 0 \\\\ \mathbf{xy+3x-3y + 1} &=& \mathbf{0} \\ \hline a&=& 3 \\ b&=& -3 \\ c &=& 1 \\ \hline \end{array} }\)

Find all pairs (x,y) of real numbers such that
\(x + y = 10 \\ x^2 + y^2 = 56 + xy.\)

\(\begin{array}{|rcll|} \hline a+b &=& 10 \\ \mathbf{b} &=& \mathbf{10-a} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline (x+y)^2 &=& x^2+2xy+y^2 \\ (x+y)^2 &=& x^2+y^2 +2xy \quad | \quad x^2 + y^2 = 56 + xy \\ 10^2 &=& 56 + xy +2xy\\ 10^2 &=& 56+3xy \\ 3xy &=& 100-56 \\ 3xy &=& 44 \quad | \quad : 3 \\ xy &=& \dfrac{44}{3} \quad | \quad y=10-x \\ x(10-x) &=& \dfrac{44}{3} \\ 10x-x^2 &=& \dfrac{44}{3} \\ \mathbf{x^2-10x+ \dfrac{44}{3}} &=& \mathbf{0} \\ \hline x &=& \dfrac{ 10\pm \sqrt{10^2 -4*\dfrac{44}{3}} }{2} \\ \\ x &=& \dfrac{ 10\pm \sqrt{4*25 -4*\dfrac{44}{3}} }{2} \\ \\ x &=& \dfrac{ 10\pm 2\sqrt{25 -\dfrac{44}{3}} }{2} \\ \\ x &=& 5\pm \sqrt{25 -\dfrac{44}{3}} \\ \\ x &=& 5\pm \sqrt{\dfrac{75-44}{3}} \\ \\ \mathbf{x} &=& \mathbf{5\pm \sqrt{\dfrac{31}{3}}} \\ \hline \end{array}\)

1.

\(\begin{array}{|rclrcl|} \hline \mathbf{x} &=& \mathbf{5+ \sqrt{\dfrac{31}{3}}} \\ &&&y &=& 10-a \\ &&&&=& 10-\left(5+ \sqrt{\dfrac{31}{3}}\right) \\ &&&\mathbf{y} &=& \mathbf{5- \sqrt{\dfrac{31}{3}}} \\ \hline \end{array}\)

2.

\(\begin{array}{|rclrcl|} \hline \mathbf{x} &=& \mathbf{5- \sqrt{\dfrac{31}{3}}} \\ &&&y &=& 10-a \\ &&&&=& 10-\left(5- \sqrt{\dfrac{31}{3}}\right) \\ &&&\mathbf{y} &=& \mathbf{5+ \sqrt{\dfrac{31}{3}}} \\ \hline \end{array}\)

Find all solutions to the system
\(a + b = 14\\ a^3 + b^3 = 812 + 3ab\)

\(\begin{array}{|rcll|} \hline a+b &=& 14 \\ \mathbf{b} &=& \mathbf{14-a} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline (a+b)^3 &=& a^3+3a^2b+3ab^2+b^3 \\ (a+b)^3 &=& a^3+b^3 +3ab(a+b) \quad | \quad a+b = 14 \\ 14^3 &=& a^3+b^3 +3*14ab \quad | \quad a^3+b^3 = 812+3ab \\ 14^3 &=& 812+3ab +3*14ab \\ 14^3 &=& 812+3*15ab \\ 3*15ab &=& 14^3- 812 \\ 3*15ab &=& 1932 \quad | \quad : 3 \\ 15ab &=& 644 \quad | \quad b=14-a \\ 15a(14-a) &=& 644 \\ a(14-a) &=& \dfrac{644}{15} \\ 14a-a^2 &=& \dfrac{644}{15} \\ \mathbf{a^2-14a+ \dfrac{644}{15}} &=& \mathbf{0} \\ \hline a &=& \dfrac{ 14\pm \sqrt{14^2 -4*\dfrac{644}{15}} }{2} \\ \\ a &=& \dfrac{ 14\pm \sqrt{4*49 -4*\dfrac{644}{15}} }{2} \\ \\ a &=& \dfrac{ 14\pm 2\sqrt{49 -\dfrac{644}{15}} }{2} \\ \\ a &=& 7\pm \sqrt{49 -\dfrac{644}{15}} \\ \\ a &=& 7\pm \sqrt{\dfrac{735-644}{15}} \\ \\ \mathbf{a} &=& \mathbf{7\pm \sqrt{\dfrac{91}{15}}} \\ \hline \end{array}\)

1.

\(\begin{array}{|rclrcl|} \hline \mathbf{a} &=& \mathbf{7+\sqrt{\dfrac{91}{15}}} \\ &&&b &=& 14-a \\ &&&&=& 14-\left(7+\sqrt{\dfrac{91}{15}}\right) \\ &&&\mathbf{b} &=& \mathbf{7-\sqrt{\dfrac{91}{15}}} \\ \hline \end{array} \)

2.

\(\begin{array}{|rclrcl|} \hline \mathbf{a} &=& \mathbf{7-\sqrt{\dfrac{91}{15}}} \\ &&&b &=& 14-a \\ &&&&=& 14-\left(7-\sqrt{\dfrac{91}{15}}\right) \\ &&&\mathbf{b} &=& \mathbf{7+\sqrt{\dfrac{91}{15}}} \\ \hline \end{array} \)

Find \(14^{-1} \pmod{35}\), as a residue modulo 35.

(Give an answer between 0 and 34, inclusive.)

Because \(\gcd(14,35) = 7 \ne 1\), so

14 is not invertible modulo 35

If \(3x+7 \equiv 11 \pmod{16}\),
then 2x+11 is congruent \(\pmod{16}\) to what integer between 0 and 15, inclusive?

\(\begin{array}{|rcll|} \hline 3x+7 &\equiv& 11 \pmod{16} \quad | \quad -7 \\ 3x &\equiv& 11-7 \pmod{16} \\ 3x &\equiv& 4 \pmod{16}\quad | \quad :3 \\ x &\equiv& \dfrac{4}{3} \pmod{16} \\ x \pmod{16} &\equiv& \dfrac{4}{3} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline 2x+11\pmod{16} &\equiv& 2*\dfrac{4}{3}+11 \pmod{16} \\ &\equiv& \dfrac{8}{3}+11 \pmod{16} \\ &\equiv& \dfrac{8}{3}+\dfrac{33}{3} \pmod{16} \\ &\equiv& \dfrac{41}{3}\pmod{16} \quad | \quad 41\equiv 9 \pmod{16} \\ &\equiv& \dfrac{9}{3}\pmod{16} \\ \mathbf{2x+11\pmod{16}} &\equiv& \mathbf{3 \pmod{16}} \\ \hline \end{array}\)

Find the sum of the first 100 terms
of the arithmetic sequence
with the first term 2
and the common difference 5.

\(\begin{array}{|rcll|} \hline \text{sum} &=& \dfrac{2+\Big( 2+(100-1)*5 \Big)}{2}*100 \\\\ \text{sum} &=& (2+2+99*5)*50 \\\\ \text{sum} &=& (4+495)*50 \\\\ \text{sum} &=& 499*50 \\\\ \mathbf{\text{sum}} &=& \mathbf{24950} \\ \hline \end{array}\)

In the diagram below,
\(\angle PQR = \angle PRQ = \angle STR = \angle TSR =\angle A,\\ RQ = 8,~ \text{ and } SQ = 2.\)
Find PQ.

\(\text{Let $\angle QPR = 180^\circ - 2A$} \\ \text{Let $\angle RQT = 180^\circ - 2A$} \\ \text{Let $\angle TRS = 180^\circ - 2A$} \\ \text{Let $\angle PTQ = 180^\circ - A$}\\ \text{Let $PQ=x$ } \\ \text{Let $QR = QT = 8$ } \\ \text{Let $TR = SR = y$ } \)

\(\begin{array}{|rcll|} \hline ST&=&QT-SQ \\ ST&=&8-2\\ ST& =& 6 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \text{In $\triangle [QPT]$:} \\ \hline \dfrac{\sin(180^\circ-2A)}{8} &=& \dfrac{\sin(180^\circ-A)}{x} \\\\ \dfrac{\sin(2A)}{8} &=& \dfrac{\sin(A)}{x} \\\\ \dfrac{\sin(2A)}{\sin(A)} &=& \dfrac{8}{x} \qquad (1)\\ \hline \end{array}\\ \begin{array}{|rcll|} \hline \text{In $\triangle [QTR]$:} \\ \hline \dfrac{\sin(180^\circ-2A)}{y} &=& \dfrac{\sin(A)}{8} \\\\ \dfrac{\sin(2A)}{y} &=& \dfrac{\sin(A)}{8} \\\\ \dfrac{\sin(2A)}{\sin(A)} &=& \dfrac{y}{8} \qquad (2)\\ \hline \end{array}\\ \begin{array}{|rcll|} \hline \text{In $\triangle [QSR]$:} \\ \hline \dfrac{\sin(180^\circ-2A)}{6} &=& \dfrac{\sin(A)}{y} \\\\ \dfrac{\sin(2A)}{6} &=& \dfrac{\sin(A)}{y} \\\\ \dfrac{\sin(2A)}{\sin(A)} &=& \dfrac{6}{y} \qquad (3)\\ \hline \end{array} \)

\(\begin{array}{|lrcll|} \hline (2)=(3):& \dfrac{y}{8} &=& \dfrac{6}{y} \\\\ & y^2 &=& 48 \\ & y^2 &=& 16*3 \\ & \mathbf{y} &=& \mathbf{4\sqrt{3}} \\ \hline \end{array} \begin{array}{|lrcll|} \hline (2)=(1):& \dfrac{y}{8} &=& \dfrac{8}{x} \\\\ & xy &=& 64 \\\\ & x &=& \dfrac{64}{y} \\\\ & x &=& \dfrac{64}{4\sqrt{3}} \\\\ & x &=& \dfrac{16}{\sqrt{3}} \\\\ & x &=& \dfrac{16}{\sqrt{3}} * \dfrac{\sqrt{3}}{\sqrt{3}} \\\\ & \mathbf{x} &=& \mathbf{\dfrac{16}{3}\sqrt{3}} \\ \hline \end{array}\)

\(PQ = \mathbf{\dfrac{16}{3}\sqrt{3}}\)