heureka

Solve 

\(\log_x (3) = \log_{81} (x)\)

\(\begin{array}{|rcll|} \hline \log_{81} (x) &=& \log_x (3) \\\\ x^{\log_{81} (x)} &=& x^{\log_x (3)} \quad | \quad x^{\log_x (3)}=3 \\\\ x^{\log_{81} (x)} &=& 3 \\ && \begin{array}{|rcll|} \hline \mathbf{x^{\log_{81} (x)}} &=& \mathbf{81^y} \quad | \quad log_{81} \text{ both sides } \\\\ log_{81}(x^{\log_{81} (x)}) &=& log_{81}(81^y) \quad | \quad log_{81}(81^y)=y \\\\ log_{81}(x^{\log_{81} (x)}) &=& y \\\\ \log_{81} (x)log_{81}(x) &=& y \\\\ \Big(\log_{81} (x)\Big)^2 &=& y \\\\ \mathbf{x^{\log_{81} (x)}} &=& \mathbf{ 81^{\Big(\log_{81} (x)\Big)^2} } \\ \hline \end{array} \\ 81^{\Big(\log_{81} (x)\Big)^2} &=& 3 \\ 3^{\color{red}4\Big(\log_{81} (x)\Big)^2} &=& 3^{\color{red}1} \\\\ \mathbf{4\Big(\log_{81} (x)\Big)^2} &=& \mathbf{1} \\\\ \Big(\log_{81} (x)\Big)^2 &=& \frac{1}{4} \\\\ \log_{81} (x) &=& \pm \sqrt{\frac{1}{4} } \\\\ \log_{81} (x) &=& \pm \frac{1}{2} \\\\ 81^{ \log_{81} (x)} &=& 81^{\pm \frac{1}{2}} \quad | \quad 81^{ \log_{81} (x)} =x \\\\ \mathbf{x} &=& \mathbf{81^{\pm \frac{1}{2}}} \\\\ x= 81^{\frac{1}{2}} &\text{or}& x= 81^{-\frac{1}{2}} \\\\ x= \sqrt{81} & & x= \dfrac{1}{\sqrt{81}} \\\\ \mathbf{x=9} & & \mathbf{ x= \dfrac{1}{9}} \\ \hline \end{array}\)

An equilateral triangle of side length 2 units is inscribed in a circle.

Find the length of a chord of this circle which passes through the midpoints of two sides of the triangle.

\(\begin{array}{|rcll|} \hline \text{In $\triangle COD$ } \\ \mathbf{r=\ ?} \\ \hline 2^2 &=& r^2+r^2-2rr\cos(120^\circ) \\ 4 &=& 2r^2\Big(1-\cos(120^\circ)\Big) \\ 2 &=& r^2\Big(1-\cos(120^\circ)\Big) \quad | \quad \cos(120^\circ) = -\dfrac{1}{2}\\ 2 &=& r^2\left(1+\dfrac{1}{2}\right) \\ 2 &=& \dfrac{3}{2}r^2 \\\\ \mathbf{r^2} &=& \mathbf{\dfrac{4}{3}} \\ \hline \end{array} \begin{array}{|rcll|} \hline \text{In $\triangle OED$ } \\ \mathbf{h=\ ?} \\ \hline 1+h^2 &=& r^2 \\ h^2 &=& r^2-1 \quad | \quad \mathbf{r^2=\dfrac{4}{3}} \\ h^2 &=& \dfrac{4}{3}-1 \\ h^2 &=& \dfrac{1}{3} \\\\ \mathbf{h^2} &=& \mathbf{\dfrac{1}{3}} \\ \mathbf{h} &=& \mathbf{\dfrac{1}{\sqrt{3}} } \\ \hline \end{array}\)

cos-rule:

\(\begin{array}{|rcll|} \hline \text{In $\triangle OEB$ } \\ \mathbf{x=\ ?} \\ \hline r^2 &=& h^2+x^2-2hx\cos(90^\circ+60^\circ) \\\\ r^2 &=& h^2+x^2-2hx\cos(150^\circ) \quad | \quad \mathbf{r^2=\dfrac{4}{3}},\ \mathbf{h^2=\dfrac{1}{3} } \\\\ \dfrac{4}{3} &=& \dfrac{1}{3}+x^2-2hx\cos(150^\circ) \quad | \quad \mathbf{h=\dfrac{1}{\sqrt{3}} },\ \cos(150^\circ)=-\dfrac{\sqrt{3}}{2} \\\\ 1 &=& x^2+\dfrac{2}{\sqrt{3}}x\dfrac{\sqrt{3}}{2} \\\\ 1 &=& x^2+x \\ \mathbf{x^2+x-1} &=& \mathbf{0} \\\\ x &=& \dfrac{-1\pm \sqrt{1^2-4(-1)} }{2} \\ x &=& \dfrac{-1\pm \sqrt{5} }{2} \\ x &=& \dfrac{-1 \mathbf{+} \sqrt{5} }{2} \\\\ \mathbf{x} &=& \mathbf{\dfrac{\sqrt{5}-1}{2}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \text{length of chord $\mathbf{= 1+2x}$} \\ \hline && 1+2x \\\\ &=& 1+ \mathbf{\dfrac{2(\sqrt{5}-1)}{2}} \\\\ &=& 1+ \sqrt{5}-1 \\ &=& \mathbf{\sqrt{5}} \\ \hline \end{array}\)

The length of the chord is \(\mathbf{\sqrt{5}} \qquad (=2.23606797750)\)

Find the area of rectangle  ABCD

\(\text{Let area of rectangle $ABCD=A$} \\ A=wh\quad \text{or}\quad \mathbf{h=\dfrac{A}{w}}\)

\(\begin{array}{|lrcll|} \hline (1) & \mathbf{\dfrac{w(h-x)}{2}} &=& \mathbf{5} \\\\ & wh-wx &=& 10 \quad | \quad wh = A \\ & A-wx &=& 10 \\ & \mathbf{wx} &=& \mathbf{A-10} \\ \hline \end{array} \begin{array}{|lrcll|} \hline (2) & \mathbf{\dfrac{h(w-y)}{2}} &=& \mathbf{8} \\\\ & wh-hy &=& 16 \quad | \quad wh = A \\ & A-hy &=& 16 \\ & \mathbf{A} &=& \mathbf{16+hy} \\ \hline \end{array} \begin{array}{|lrcll|} \hline (3) & \mathbf{\dfrac{xy}{2}} &=& \mathbf{9} \\\\ & xy &=& 18 \\ & \mathbf{y} &=& \mathbf{\dfrac{18}{x}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{A} &=& \mathbf{16+hy} \quad | \quad \mathbf{h=\dfrac{A}{w}},\ \mathbf{y=\dfrac{18}{x}} \\\\ A &=& 16+\dfrac{A}{w}*\dfrac{18}{x} \\\\ A &=& 16+\dfrac{18A}{wx} \\\\ A-16 &=& \dfrac{18A}{wx} \quad | \quad \mathbf{wx=A-10} \\\\ A-16 &=& \dfrac{18A}{A-10} \\\\ (A-16)(A-10) &=& 18A \\ A^2-26A+160 &=& 18A \\ \mathbf{A^2-44A+160} &=& \mathbf{0} \\\\ A &=& \dfrac{44\pm \sqrt{44^2-4*160}} {2} \\\\ A &=& \dfrac{44\pm \sqrt{1296}} {2} \\\\ A &=& \dfrac{44\pm 36} {2} \\\\ A = \dfrac{44+ 36} {2} &\text{or}& A = \dfrac{44- 36} {2} \\\\ A = \dfrac{80} {2} && A = \dfrac{8} {2} \\\\ \mathbf{A = 40} && A = 4 \quad \text{no solution!} \\ \hline \end{array}\)

The area of rectangle  ABCD is \(\mathbf{40}\)

Simplify

\(\dfrac { { a }^{ 3 }(b+c) }{ (a-b)(a-c) } +\dfrac { { b }^{ 3 }(a+c) }{ (b-a)(b-c) } +\dfrac { { c }^{ 3 }(b+a) }{ (c-b)(c-a) }\)

\(\begin{array}{|rcll|} \hline && \mathbf{\dfrac {a^3(b+c) }{ (a-b)(a-c) } +\dfrac {b^3(a+c)}{ (b-a)(b-c) } +\dfrac {c^3(b+a)}{ (c-b)(c-a) } } \\\\ &=& \dfrac {a^3(b+c) }{ (a-b)(a-c) } -\dfrac {b^3(a+c)}{ (a-b)(b-c) } +\dfrac {c^3(a+b)}{ (b-c)(a-c) } \\\\ &=& \dfrac {a^3(b+c)(b-c)- b^3(a+c)(a-c)+c^3(a+b)(a-b)}{ (a-b)(a-c)(b-c) } \\\\ &=& \dfrac {a^3(b+c)(b-c)+c^3(a+b)(a-b)- b^3(a+c)(a-c)}{ (a-b)(a-c)(b-c) } \\\\ &=& \dfrac {a^3(b^2-c^2)+c^3(a^2-b^2)- b^3(a+c)(a-c)}{ (a-b)(a-c)(b-c) } \\\\ &=& \dfrac {a^3b^2-a^3c^2+c^3a^2-c^3b^2- b^3(a+c)(a-c)}{ (a-b)(a-c)(b-c) } \\\\ &=& \dfrac {a^3b^2-c^3b^2-a^3c^2+c^3a^2- b^3(a+c)(a-c)}{ (a-b)(a-c)(b-c) } \\\\ &=& \dfrac {b^2(a^3-c^3)-a^2c^2(a-c)- b^3(a+c)(a-c)}{ (a-b)(a-c)(b-c) } \\\\ &=& \dfrac {b^2(a-c)(a^2+ac+c^2)-a^2c^2(a-c)- b^3(a+c)(a-c)}{ (a-b)(a-c)(b-c) } \\\\ &=& \dfrac {(a-c)\left(b^2(a^2+ac+c^2)-a^2c^2- b^3(a+c)\right)}{ (a-b)(a-c)(b-c) } \\\\ &=& \dfrac { b^2(a^2+ac+c^2)-a^2c^2- b^3(a+c) }{ (a-b)(b-c) } \\\\ &=& \dfrac { b^2a^2+b^2ac+b^2c^2 -a^2c^2- b^3a-b^3c }{ (a-b)(b-c) } \\\\ &=& \dfrac { b^2a^2- b^3a +b^2ac-b^3c +b^2c^2 -a^2c^2 }{ (a-b)(b-c) } \\\\ &=& \dfrac { ab^2(a- b) +b^2c(a-b) -c^2( a^2-b^2) }{ (a-b)(b-c) } \\\\ &=& \dfrac { (ab^2+b^2c)(a- b)-c^2(a+b)(a-b) }{ (a-b)(b-c) } \\\\ &=& \dfrac {(a-b)\left( (ab^2+b^2c)-c^2(a+b)\right) }{ (a-b)(b-c) } \\\\ &=& \dfrac {ab^2+b^2c -c^2(a+b)}{(b-c) } \\\\ &=& \dfrac {ab^2+b^2c -c^2a-c^2b}{(b-c) } \\\\ &=& \dfrac {ab^2 -c^2a +b^2c-c^2b}{(b-c) } \\\\ &=& \dfrac {a(b^2 -c^2) +bc(b-c)}{(b-c) } \\\\ &=& \dfrac {a(b+c)(b-c) +bc(b-c)}{(b-c) } \\\\ &=& \dfrac {(b-c)\Big(a(b+c) +bc\Big)}{(b-c) } \\\\ &=& a(b+c) +bc \\ &=& \mathbf{ab+ac+bc} \\ \hline \end{array}\)

We are given a triangle with a perimeter of 272 and the product of its sides equal to 314.  
What is the product of the inradius and circumradius?

\(\text{Let inradius $r = \dfrac{2A}{a+b+c} $ } \\ \text{Let circumradius $ R =\dfrac{abc}{4A} $ } \\ a+b+c = 272 \\ abc = 314 \)

\(\begin{array}{|rcll|} \hline rR &=& \dfrac{2A}{(a+b+c)} * \dfrac{abc}{4A} \\\\ rR &=& \dfrac{abc}{2(a+b+c)} \\\\ rR &=& \dfrac{314}{2(272)} \\\\ rR &=& \dfrac{157}{272} \\\\ \mathbf{rR} &=& \mathbf{0.57720588235} \\ \hline \end{array}\)

In the sequence \(1,\ 4,\ 4,\ 9,\ 9,\ 9,\ 16,\ 16,\ 16,\ 16, \ \ldots\)
Find the sum of first 99 terms.

\(\begin{array}{|rcll|} \hline && 1,\ 4,\ 4,\ 9,\ 9,\ 9,\ 16,\ 16,\ 16,\ 16, \ \ldots ,\ 169,\ 196,\ 196,\ 196,\ 196,\ 196,\ 196,\ 196,\ 196 \\ s &=& 1*1^2+2*2^2+3*3^2+4*4^2+5*5^2+6*6^2+ \ldots + 13*13^2 + 8*14^2 \\ s &=& 1^3+2^3+3^3+4^3+5^3+6^3+7^3+8^3+9^3+10^3+11^3+12^3+ 13^3 + 8*14^2 \\ s &=& \left(\dfrac{13*(13+1)}{2}\right)^2+ 8*14^2 \\ s &=& \left(\dfrac{13*14}{2}\right)^2+1568 \\ s &=& (13*7)^2+1568 \\ s &=& 91^2+1568 \\ s &=& 8281+1568 \\ \mathbf{s} &=& \mathbf{9849} \\ \hline \end{array}\)

TP is a line that tangent to a circle centered at \(O\).
If \(PQ\parallel TO\) and \(\angle OTP=25^\circ\), find the measure of \(\angle POQ\) in degrees.

\(\begin{array}{|rcll|} \hline \hline \mathbf{65^\circ} &=& \mathbf{90^\circ-\dfrac{x}{2}} \\\\ \dfrac{x}{2} &=& 90^\circ-65^\circ \\\\ \dfrac{x}{2} &=& 25^\circ \\\\ \mathbf{x} &=& \mathbf{50^\circ} \\ \end{array}\)

Find the Relationship

\(\mathbf{A=i+{\dfrac {p}{2}}-1}.\)

Angle bisectors \(\overline{AX}\) and \(\overline{BY}\) of triangle \(ABC\) meet at point \(I\).
Find \(\angle C\), in degrees, if \(\angle AIB = 109^\circ\).

\(\begin{array}{|rcll|} \hline \text{In }\triangle AIB \\ \hline \mathbf{180^\circ} &=& \mathbf{109^\circ + A + B} \\ 180^\circ-109^\circ &=& A + B \\ 71^\circ &=& A + B \\ \mathbf{A+B} &=& \mathbf{71^\circ} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \text{In } AIXC \\ \hline \mathbf{360^\circ} &=& \mathbf{C + 109^\circ + 180^\circ-(C+B) + 180^\circ-(C+A)} \\ 0 &=& C + 109^\circ -(C+B) -(C+A) \\ 0 &=& C + 109^\circ -2C - (A+B) \\ 0 &=& -C + 109^\circ - (A+B) \\ C &=& 109^\circ - (A+B) \quad | \quad \mathbf{A+B=71^\circ} \\ C &=& 109^\circ - 71^\circ \\ \mathbf{C} &=& \mathbf{38^\circ} \\ \hline \end{array}\)

Let \(\triangle MBT\) be a triangle with MB = 4 and MT = 7.
Furthermore, let circle \(\omega\) be a circle with center O which is tangent to MB at B and MT at some point on segment MT.
Given OM = 6 and \(\omega\) intersect BT at \(I \ne B\), find the length of \(TI\).

\(\text{Let $TI = \color{red}x$}\)

\(\begin{array}{|rcll|} \hline \mathbf{r=\ ?} \\ \hline 4^2+r^2 &=& 6^2 \\ r^2 &=& 6^2 -4^2 \\ r^2 &=& 36-16 \\ r^2 &=& 20 \\ r^2 &=& 4*5 \\ \mathbf{r} &=& \mathbf{2\sqrt{5}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \cos(A) &=& \dfrac{4}{6} \\\\ \cos(A) &=& \dfrac{2}{3} \\\\ \mathbf{\cos^2(A)} &=& \mathbf{\dfrac{4}{9}} \\\\ \cos(2A) &=& \cos^2(A)-\sin^2(a) \\ \cos(2A) &=& \cos^2(A)-\left(1- cos^2(A) \right) \\ \mathbf{\cos(2A)} &=& \mathbf{2\cos^2(A)-1} \\\\ \cos(2A) &=& 2*\dfrac{4}{9}-1 \\\\ \cos(2A) &=& \dfrac{8}{9}-1 \\\\ \cos(2A) &=& \dfrac{8}{9}-\dfrac{9}{9} \\\\ \mathbf{\cos(2A)} &=& \mathbf{-\dfrac{1}{9}} \\ \hline \end{array}\)

cos-rule:

\(\begin{array}{|rcll|} \hline BT^2 &=& 4^2+7^2-2*4*7*\cos(2A) \\ BT^2 &=& 16+49-56\cos(2A) \quad | \quad \mathbf{\cos(2A)=-\dfrac{1}{9}} \\ BT^2 &=& 16+49+56*\dfrac{1}{9} \\ BT^2 &=& 65+\dfrac{56}{9} \\ BT^2 &=& \dfrac{9*65+56}{9} \\ BT^2 &=& \dfrac{641}{9} \\ \mathbf{BT} &=& \mathbf{\dfrac{\sqrt{641}}{3}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{x=\ ?} \\ \hline 3^2 &=& x(x+BI) \quad | \quad BI = BT -x \\ 9 &=& x(x+BT -x) \\ 9 &=& x*BT \\ x &=& \dfrac{9}{BT} \quad | \quad \mathbf{BT=\dfrac{\sqrt{641}}{3}} \\\\ x &=& \dfrac{9}{\dfrac{\sqrt{641}}{3}} \\\\ x &=& \dfrac{3*9}{ \sqrt{641} } \\\\ x &=& \dfrac{27}{ \sqrt{641} } \\\\ x &=& \dfrac{27}{ 25.3179778023 } \\\\ \mathbf{x} &=& \mathbf{1.06643588247} \\ \hline \end{array}\)

The length of TI is  \(\approx \mathbf{1}\)