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Triangle ABC has vertices B and C on a semicircle centered at O, as shown, with AB tangent to the semicircle at B and AC intersecting the semicircle at point D.

If angle BAC = 78 degrees (in red) and angle ODC = 74  degrees (in green),

what is the measure of angle BCA (in blue) in degrees?

\(\begin{array}{|rcll|} \hline \mathbf{x+\varphi} &=& \mathbf{74^\circ} \quad | \quad \text{in triangle ODC} \\ \mathbf{\varphi} &=& \mathbf{74^\circ-x } \\\\ \mathbf{ 78^\circ +(90^\circ-\varphi) +x} &=& \mathbf{180^\circ} \quad | \quad \text{in triangle BAC} \\ 78^\circ +\Big(90^\circ-(74^\circ-x)\Big) +x &=& 180^\circ \\ 78^\circ +(90^\circ-74^\circ+x) +x &=& 180^\circ \\ 78^\circ + 90^\circ-74^\circ+x +x &=& 180^\circ \\ 94^\circ+2x &=& 180^\circ \quad | \quad :2 \\ 47^\circ+ x &=& 90^\circ \\ x &=& 90^\circ -47^\circ \\ \mathbf{x} &=& \mathbf{43^\circ} \\ \hline \end{array}\)

The measure of angle BCA is \(\mathbf{43^\circ}\)

If

\(\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{16} + \dfrac{1}{n} + \dfrac{1}{62} + \dfrac{1}{126} + \dfrac{1}{248} = \dfrac{495}{496}\)

 what is the value of \(n\)?

\(\begin{array}{|rcll|} \hline \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{n} + \frac{1}{62} + \frac{1}{126} + \frac{1}{248} &=& \frac{495}{496} \quad | \quad \times 2 \\ \frac{2}{2} + \frac{2}{4} + \frac{2}{8} + \frac{2}{16} + \frac{2}{n} + \frac{2}{62} + \frac{2}{126} + \frac{2}{248} &=& \frac{495}{248} \\ 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{2}{n} + \frac{1}{31} + \frac{1}{63} + \frac{2}{248} &=& \frac{495}{248} \\ 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{2}{n} + \frac{1}{31} + \frac{1}{63} &=& \frac{495}{248} - \frac{2}{248} \\ 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{2}{n} + \frac{1}{31} + \frac{1}{63} &=& \frac{493}{248} \quad | \quad \times 8 \\ 8 + \frac{8}{2} + \frac{8}{4} + \frac{8}{8} + \frac{16}{n} + \frac{8}{31} + \frac{8}{63} &=& \frac{493}{31} \\ 8 + 4 + 2 + 1 + \frac{16}{n} + \frac{8}{31} + \frac{8}{63} &=& \frac{493}{31} \\ 15 + \frac{16}{n} + \frac{8}{63} &=& \frac{493}{31} - \frac{8}{31} \\ 15 + \frac{16}{n} + \frac{8}{63} &=& \frac{485}{31} \\ \frac{16}{n} + \frac{8}{63} &=& \frac{485}{31} -15 \\ \frac{16}{n} + \frac{8}{63} &=& \frac{485-15*31}{31} \\ \frac{16}{n} + \frac{8}{63} &=& \frac{20}{31} \\ \frac{16}{n} &=& \frac{20}{31} - \frac{8}{63} \\ \frac{16}{n} &=& \frac{20*63-8*31}{31*63} \\ \frac{16}{n} &=& \frac{1012}{1953} \\ \frac{1}{n} &=& \frac{1012}{16*1953} \\ \frac{1}{n} &=& \frac{1012}{31248} \\ \frac{1}{n} &=& \frac{253}{7812} \\ \mathbf{n} &=& \mathbf{\frac{7812}{253}} \\ \hline \end{array}\)

In triangle ABC, E is on AB such that AE:EB = 3:2, and D is on AC such that AD:DC = 1:4.  
F is the intersection of BD and CE.  
Find EF:FC.

\(\begin{array}{lclcll} \text{Let $\vec{AB}=\vec{b}$ } && \text{Let $\vec{AC}=\vec{c}$ }\\\\ \text{Let $\dfrac{AE}{AB}=\lambda=\dfrac{3}{5} $ } && \text{Let $1-\lambda=\dfrac{2}{5} $ }\\ \text{Let $\dfrac{AD}{AC}=\mu=\dfrac{1}{5} $ } && \text{Let $1-\mu=\dfrac{4}{5} $ } \\ \text{Let $\dfrac{EF}{EC}=x $ } && \text{Let $\dfrac{FC}{EC}=1-x $ }\\ \text{Let $\dfrac{EF}{FC}=\frac{x}{1-x} $ }\\\\ \text{Let $\dfrac{DF}{DB}=\eta $ }\\\\ \text{Let $\vec{AE}=\lambda\vec{b}$ } && \text{Let $\vec{AD}=\mu\vec{c}$ }\\\\ \text{Let $\vec{EC}=\vec{AC}-\vec{AE}=\vec{c}-\lambda\vec{b}$ } &&\text{Let $\vec{DB}=\vec{AB}-\vec{AD}=\vec{b}-\mu\vec{c}$ } \\\\ \text{Let $\vec{EF}=x\vec{EC}=x(\vec{c}-\lambda\vec{b})$ } &&\text{Let $\vec{DF}=\eta\vec{DB}=\eta(\vec{b}-\mu\vec{c})$ } \\ \end{array} \)

\(\begin{array}{|rcll|} \hline \vec{AE} + \vec{EF} &=& \vec{AD} + \vec{DF} \\ \lambda\vec{b} + x(\vec{c}-\lambda\vec{b}) &=& \mu\vec{c} + \eta(\vec{b}-\mu\vec{c}) \\ \lambda\vec{b} + x\vec{c}-x\lambda\vec{b} &=& \mu\vec{c} + \eta\vec{b}-\eta\mu\vec{c} \\ \vec{b}(\underbrace{\lambda-x\lambda- \eta}_{=0} ) &=& \vec{c} (\underbrace{\mu-\eta\mu- x}_{=0}) \\ \hline \lambda-x\lambda- \eta &=& 0 \\ \eta &=& \lambda-x\lambda \\ \mathbf{ \eta } &=& \mathbf{ \lambda(1-x) } \\ \hline \mu-\eta\mu- x &=& 0 \\ x &=& \mu-\eta\mu \\ x &=& \mu-\lambda(1-x)\mu \\ x &=& \mu-\lambda\mu(1-x) \\ x &=& \mu-\lambda\mu + \lambda\mu x \\ x - \lambda\mu x &=& \mu-\lambda\mu \\ x (1- \lambda\mu) &=& \mu(1-\lambda) \\ \mathbf{x} &=& \mathbf{\dfrac{ \mu(1-\lambda)}{1- \lambda\mu} } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{x} &=& \mathbf{\dfrac{ \mu(1-\lambda)}{1- \lambda\mu} } \\\\ x &=& \dfrac{ \dfrac{1}{5}* \dfrac{2}{5}}{1- \dfrac{3}{5}*\dfrac{1}{5} } \\\\ x &=& \dfrac{ \dfrac{2}{25}}{1- \dfrac{3}{25} } \\\\ x &=& \dfrac{ \dfrac{2}{25}}{ \dfrac{22}{25} } \\\\ x &=& \dfrac{2}{22} \\\\ \mathbf{x} &=& \mathbf{\dfrac{1}{11}} \\\\ \mathbf{1-x} = 1-\dfrac{1}{11} &=& \mathbf{\dfrac{10}{11}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \dfrac{EF}{FC} &=& \dfrac{x}{1-x} \\\\ \dfrac{EF}{FC} &=&\dfrac{ \dfrac{1}{11} } { \dfrac{10}{11} } \\\\ \mathbf{\dfrac{EF}{FC}} &=& \mathbf{\dfrac{1}{10}} \\ \hline \end{array}\)

Let PQRS be a square with side length 10.
The points X and Y lie outside the square such that XQ = 6, XP = 8, YS = 6, and YR = 8.
Find \(XY^2\).

\(\begin{array}{|rcll|} \hline XY^2 &=& (6+8)^2 + (6+8)^2 \\ XY^2 &=& 2(6+8)^2 \\ XY^2 &=& 2(14)^2 \\ \mathbf{XY^2} &=& \mathbf{392} \\ \hline \end{array}\)

Let B, A, and D be three consecutive vertices of a regular 18-gon.
A regular heptagon is constructed on \(\overline{AB}\), with a vertex C next to A.
Find \(\angle BCD\), in degrees.

The inscribed \(\angle BCD\) is half of the central \(\angle BAD\)

\(\begin{array}{|rcll|} \hline \angle BCD &=& \dfrac{\angle BAD}{2} \\ && \boxed{\angle BAD = \dfrac{16*180^\circ}{18}\\ \angle BAD =160^\circ } \\ \angle BCD &=& \dfrac{160^\circ}{2} \\ \mathbf{\angle BCD} &=& \mathbf{80^\circ } \\ \hline \end{array}\)

The quadrilateral ABCD has right angles at A and D.

The numbers show the areas of the colored triangles.

Find the area of ABC

\(\begin{array}{|rcll|} \hline \dfrac{hb}{2} &=& 16 + 48 \quad \text{or} \quad \mathbf{\dfrac{hb}{2}=64} \qquad (1) \\\\ \mathbf{\dfrac{hb}{2}} &=& \mathbf{16+A_1} \qquad (2) \\ \hline \dfrac{hb}{2} = 64 &=& 16+A_1 \\ 64 &=& 16+A_1 \\ A_1 &=& 64-16 \\ \mathbf{A_1} &=& \mathbf{48} \\ \hline \end{array} \begin{array}{|rcll|} \hline \dfrac{ha}{2} &=& A_1+A_2 \quad |\quad \mathbf{A_1=48} \\\\ \mathbf{\dfrac{ha}{2}} &=& \mathbf{48+A_2} \qquad (3) \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \dfrac{bh_1}{2} &=& 16 \quad \text{or} \quad \mathbf{h_1=\dfrac{32}{b}} \qquad (4) \\ \hline \dfrac{ah_2}{2} &=& A_2 \quad | \quad h = h_1+h_2 \quad \text{or} \quad \mathbf{h_2=h-h_1} \\\\ \dfrac{a(h-h_1)}{2} &=& A_2 \\\\ \dfrac{ah}{2} - \dfrac{ah_1}{2} &=& A_2\quad | \quad \mathbf{h_1=\dfrac{32}{b}} \\\\ \dfrac{ah}{2} - \dfrac{a\dfrac{32}{b}}{2} &=& A_2 \\\\ \dfrac{ah}{2} - \dfrac{16a}{b} &=& A_2 \quad | \quad \mathbf{\dfrac{ha}{2}= 48+A_2} \\\\ 48+A_2 - \dfrac{16a}{b} &=& A_2 \\\\ 48 - \dfrac{16a}{b} &=& 0 \\\\ \dfrac{16a}{b} &=& 48 \\\\ \dfrac{a}{b} &=& \dfrac{48}{16} \\\\ \mathbf{\dfrac{a}{b}} &=& \mathbf{3} \qquad (5) \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline \dfrac{(3)}{(1)}: & \dfrac{ \dfrac{ah}{2} } { \dfrac{bh}{2} } &=& \dfrac{48+A_2}{64} \\\\ & \dfrac{ah}{2} \cdot \dfrac{2}{bh} &=& \dfrac{48+A_2}{64} \\\\ & \dfrac{a}{b} &=& \dfrac{48+A_2}{64} \quad | \quad \mathbf{\dfrac{a}{b}=3} \\\\ & 3 &=& \dfrac{48+A_2}{64} \\\\ & 192 &=& 48+A_2 \\ & A_2 &=& 192-48 \\ & \mathbf{A_2} &=& \mathbf{144} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \text{The area of $ABCD$} &=& 16+48+A_1+A_2 \\ &=& 16+48+48+144 \\ \mathbf{\text{The area of $ABCD$}} &=& \mathbf{256} \\ \hline \end{array} \)

How many 9-step paths are there from E to G?

Second soultion:

How many 9-step paths are there from E to G?

\(\begin{array}{|rcll|} \hline \text{Permutations of (hhhhhvvvv)} \\ \hline && \dfrac{(5+4)!}{5!4!} \\\\ &=& \dfrac{9!}{5!4!} \\\\ &=& \dfrac{5!6*7*8*9}{5!4!} \\\\ &=& \dfrac{6*7*8*9}{4!} \\\\ &=& \dfrac{6*7*8*9}{2*3*4} \\\\ &=& \dfrac{6*7*8*9}{2*3*4} \\\\ &=& \dfrac{3024}{24} \\\\ &=& 126 \\ \hline \end{array}\)

Find the number of integer solutions to \(a^2 + b^2 = 625\).

\(\begin{array}{|rclcrcl|} \hline a^2 + b^2 &=& 625& \text{or}& \mathbf{a^2 + b^2} &=& \mathbf{25^2} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \mathbf{3^2+4^2} &=& \mathbf{5^2} \\\\ (3*5)^2+(4*5)^2 &=& (5*5)^2 \\ \mathbf{15^2+20^2} &=& \mathbf{25^2} \\ \hline \end{array}\)

\(a=15~ \text{and}~ b= 20 \)

There are other solutions for example: \(7^2+24^2 = 25^2\quad \text{ (See: Pythagorean triple )}\)

Four positive integers, A, B, C, D have a sum of 36.

If A+2=B-2=C*2=D/2.

What is A*B*C*D?

\(\begin{array}{|rcll|} \hline A+2 &=& B-2 \\ \mathbf{A} &=& \mathbf{B-4} \\ \hline \end{array} \begin{array}{|rcll|} \hline 2C &=& B-2 \\ \mathbf{C} &=& \mathbf{\dfrac{B-2}{2}} \\ \hline \end{array} \begin{array}{|rcll|} \hline \dfrac{D}{2} &=& B-2 \\ \mathbf{D} &=& \mathbf{2(B-2)} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline A+B+C+D &=& 36 \\ (B-4)+B+\dfrac{B-2}{2}+2(B-2) &=& 36 \\ B-4+B+\dfrac{B}{2}-1+2B-4 &=& 36 \\ 4B+\dfrac{B}{2}-9 &=& 36 \\ 4B+\dfrac{B}{2} &=& 36+9 \\ 4B+\dfrac{B}{2} &=& 45 \quad | \quad \cdot 2 \\ 8B+B &=& 90 \\ 9B &=& 90 \quad | \quad : 9 \\ \mathbf{B} &=& \mathbf{10} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{A} &=& \mathbf{B-4} \\ A &=& 10-4 \\ \mathbf{A} &=& \mathbf{6} \\ \hline \end{array} \begin{array}{|rcll|} \hline \mathbf{C} &=& \mathbf{\dfrac{B-2}{2}} \\ C &=& \dfrac{10-2}{2} \\ \mathbf{C} &=& \mathbf{4} \\ \hline \end{array} \begin{array}{|rcll|} \hline \mathbf{D} &=& \mathbf{2(B-2)} \\ D &=& 2(10-2) \\ \mathbf{D} &=& \mathbf{16} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline A*B*C*D &=& 6*10*4*16 \\ \mathbf{A*B*C*D}&=&\mathbf{3840} \\ \hline \end{array}\)