heureka

AB is the diameter of the semicircle.
C is a point on the circumference.
O is the center if the semicircle.

If the radius of the semicircle is 5, \(\angle COA=60^\circ\).
Find the sum of the area of the region R and S.

Note region R and S is not the semicircle with the diameter AC and BC

\(\text{Let segment of the circle $AC = A_1$} \\ \text{Let segment of the circle $CB = A_2$}\)

\(\begin{array}{|rcll|} \hline \mathbf{A_1 + A_2} &=& \mathbf{\text{semicircle AB} - \triangle ABC} \\\\ R+S &=& (\text{semicircle AC}-A_1) + (\text{semicircle CB}-A_2) \\ R+S &=& \text{semicircle AC}+\text{semicircle CB}-(A_1+A_2) \\ R+S &=& \text{semicircle AC}+\text{semicircle CB}-(\text{semicircle AB} - \triangle ABC) \\ R+S &=& \text{semicircle AC}+\text{semicircle CB}-\text{semicircle AB} + \triangle ABC \\ R+S &=& \dfrac{\pi \left(\dfrac{5}{2}\right)^2 }{2} +\dfrac{\pi \left(\dfrac{5\sqrt{3}}{2}\right)^2 }{2}-\dfrac{\pi 5^2 }{2} + \triangle ABC \\\\ R+S &=& \dfrac{5^2}{8}\pi +\dfrac{3*5^2}{8}\pi -\dfrac{5^2}{2}\pi + \triangle ABC \\\\ R+S &=& \dfrac{5^2}{8}\pi +\dfrac{3*5^2}{8}\pi -\dfrac{4*5^2}{8}\pi + \triangle ABC \\\\ R+S &=& \dfrac{4*5^2}{8}\pi-\dfrac{4*5^2}{8}\pi + \triangle ABC \\\\ \mathbf{R+S} &=& \mathbf{\triangle ABC} \\ \hline R+S &=& \dfrac{5*5\sqrt{3}}{2} \\\\ \mathbf{ R+S } &=& \mathbf{\dfrac{25\sqrt{3}}{2}} \\ \hline \end{array}\)

The figure shows a square ABCD of side 6.
It contains arcs BD and AC drawn with centers at A and D, respectively.
If x is the radius of the circle tangent to AB and arcs AC and BD. What is the value of x?

\(\begin{array}{|lrcll|} \hline (1) & (6+x)^2 = (6-x)^2 + y^2 \\ (2) & (6-x)^2 = x^2 + y^2 \\ \hline (1)-(2): & (6+x)^2-(6-x)^2 &=& (6-x)^2-x^2 \\ & (6+x)^2 &=& (6-x)^2+(6-x)^2-x^2 \\ & (6+x)^2 &=& 2(6-x)^2-x^2 \\ & 36+12x+x^2 &=& 2(36-12x+x^2) -x^2 \\ & 36+12x+x^2 &=& 2*36-24x+2x^2 -x^2 \\ & 36+12x &=& 2*36-24x \\ & 36+36x &=& 2*36 \\ & 36x &=& 36 \\ & \mathbf{x} &=& \mathbf{1} \\ \hline \end{array}\)

Lines \(XTQ\) and \(XUR\) are tangent to a circle, as shown below.

If \(\angle ATQ = 41^\circ\) and \(\angle AUR = 63^\circ\), then find \(\angle QXR\), in degrees.

\(\text{Let $\angle QXR = x $ }\)

\(\angle TCX = 360^\circ -2*90^\circ = 180^\circ - x\)

\(\begin{array}{|rcll|} \hline \mathbf{360^\circ} &=& \mathbf{(180^\circ - x)+82^\circ+126^\circ} \\ 360^\circ &=& 388^\circ - x \\ x &=& 388^\circ - 360^\circ \\ \mathbf{x} &=& \mathbf{28^\circ} \\ \hline \end{array}\)

The \(\angle QXR\) is \(\mathbf{28^\circ}\)

The legs of a right triangle are 8 and 15.  Find the altitude to the hypotenuse.

Formula: \( h_c = \dfrac{ab}{c}\) (right triangle)

\(\begin{array}{|rcll|} \hline \mathbf{h_c} &=& \mathbf{\dfrac{ab}{c}} \quad | \quad c = \sqrt{a^2+b^2} \\\\ h_c &=& \dfrac{ab}{\sqrt{a^2+b^2}} \\\\ h_c &=& \dfrac{8*15}{\sqrt{8^2+15^2}} \\\\ h_c &=& \dfrac{120}{\sqrt{289}} \\\\ \mathbf{h_c} &=& \mathbf{\dfrac{120}{17}} \\ \hline \end{array}\)

Solve

\(\cos^7(x) + \cos^7\left(x+\dfrac{2\pi}{3}\right) + \cos^7\left(x+\dfrac{4\pi}{3}\right)=0 \) for \(0\leq x\leq 2\pi\)

Formula:

\(\begin{array}{|rcll|} \hline \cos^{n}(x) &=& {\frac {1}{2^{n}}}\,\sum \limits_{k=0}^{n}{n \choose k}\,\cos\Big((n-2k)x\Big);\quad n\in \mathbb {N} \\\\ \cos^{7}(x) &=& {\frac {1}{2^{7}}}\,\sum \limits_{k=0}^{7}{7 \choose k}\,\cos\Big((7-2k)x\Big) \\ \hline \mathbf{\cos ^{7}(x)} &=& \mathbf{{ \frac {1}{2^{7}} } \Big( \cos(7x)+7\cos(5x) +21\cos(3x)+35\cos(x) \Big)} \\\\ \cos^{7}\left(x+\dfrac{2\pi}{3}\right) &=& { \frac {1}{2^{7}} } \Bigg( \cos(7\left(x+\dfrac{2\pi}{3}\right))+7\cos(5\left(x+\dfrac{2\pi}{3}\right)) +21\cos(3\left(x+\dfrac{2\pi}{3}\right))+35\cos\left(x+\dfrac{2\pi}{3}\right) \Bigg) \\ &=& { \frac {1}{2^{7}} } \Bigg( \cos\left(7x+\dfrac{14\pi}{3}\right) +7\cos\left(5x+\dfrac{10\pi}{3}\right) +21\cos(3x+2\pi)+35\cos\left(x+\dfrac{2\pi}{3}\right) \Bigg) \\ \mathbf{\cos^{7}\left(x+\dfrac{2\pi}{3}\right)} &=& \mathbf{{ \frac {1}{2^{7}} } \Bigg( \cos\left(7x+\dfrac{14\pi}{3}\right) +7\cos\left(5x+\dfrac{10\pi}{3}\right) +21\cos(3x)+35\cos\left(x+\dfrac{2\pi}{3}\right) \Bigg) } \\\\ \cos^{7}\left(x+\dfrac{4\pi}{3}\right) &=& { \frac {1}{2^{7}} } \Big( \cos(7\left(x+\dfrac{4\pi}{3}\right))+7\cos(5\left(x+\dfrac{4\pi}{3}\right)) +21\cos(3\left(x+\dfrac{4\pi}{3}\right))+35\cos\left(x+\dfrac{4\pi}{3}\right) \Bigg) \\ &=& { \frac {1}{2^{7}} } \Bigg( \cos\left(7x+\dfrac{28\pi}{3}\right) +7\cos\left(5x+\dfrac{20\pi}{3}\right) +21\cos(3x+4\pi)+35\cos\left(x+\dfrac{4\pi}{3}\right) \Bigg) \\ \mathbf{\cos^{7}\left(x+\dfrac{4\pi}{3}\right)} &=& \mathbf{{ \frac {1}{2^{7}} } \Bigg( \cos\left(7x+\dfrac{28\pi}{3}\right) +7\cos\left(5x+\dfrac{20\pi}{3}\right) +21\cos(3x)+35\cos\left(x+\dfrac{4\pi}{3}\right) \Bigg) } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && \cos^7(x) + \cos^7\left(x+\dfrac{2\pi}{3}\right) + \cos^7\left(x+\dfrac{4\pi}{3}\right) \\ &=& \frac {1}{128} \Bigg( 61\cos(3x) \\ && + \cos(7x) + \cos\left(7x+\dfrac{14\pi}{3}\right) + \cos\left(7x+\dfrac{28\pi}{3}\right)\\ && + 7\Big( \cos(5x) + \cos\left(5x+\dfrac{10\pi}{3}\right) + \cos\left(5x+\dfrac{20\pi}{3}\right) \Big) \\ && +35\Big( \cos(x) + \cos\left(x+\dfrac{2\pi}{3}\right) + \cos\left(x+\dfrac{4\pi}{3}\right) \Big) \Bigg) \\ \hline \end{array} \)

\(\text{Formula}: \cos x+\cos y=2\cos \frac{x+y}{2}\cos \frac{x-y}{2} \)

\(\begin{array}{|rcll|} \hline && \mathbf{\cos(7x) + \cos\left(7x+\dfrac{14\pi}{3}\right) + \cos\left(7x+\dfrac{28\pi}{3}\right)} \\\\ &=& \cos(7x) + 2\cos\left(\dfrac{14x+14\pi}{2}\right) \cos\left(\dfrac{\dfrac{14\pi}{3}}{2}\right) \\\\ &=& \cos(7x) + 2\cos(7x+7\pi) \cos\left(\dfrac{7\pi}{3}\right) \\\\ &=& \cos(7x) + 2\cos(7x+\pi) \cos\left(\dfrac{7\pi}{3}\right) \\\\ &=& \cos(7x) + 2\Big( \cos(7x)\cos(\pi)-\sin(7x)\sin(\pi) \Big) \cos\left(\dfrac{7\pi}{3}\right) \\\\ &=& \cos(7x) - 2\cos(7x) \cos\left(\dfrac{7\pi}{3}\right) \\\\ &=& \cos(7x) \Big(1 - 2\cos\left(\dfrac{7\pi}{3}\right)\Big) \quad | \quad \cos\left(\dfrac{7\pi}{3}\right)=\dfrac{1}{2} \\\\ &=& \cos(7x) (1 - 2*\dfrac{1}{2}) \\\\ &=& \cos(7x) (1 - 1) \\\\ &=& \mathbf{ 0 } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && \mathbf{7\Bigg( \cos(5x) + \cos\left(5x+\dfrac{10\pi}{3}\right) + \cos\left(5x+\dfrac{20\pi}{3}\right) \Bigg)} \\\\ &=& 7\Bigg(\cos(5x) + 2\cos\left(\dfrac{10x+10\pi}{2}\right) \cos\left(\dfrac{\dfrac{10\pi}{3}}{2}\right) \Bigg)\\\\ &=& 7\Bigg(\cos(5x) + 2\cos(5x+5\pi) \cos\left(\dfrac{5\pi}{3}\right) \Bigg)\\\\ &=& 7\Bigg(\cos(5x) + 2\cos(5x+\pi) \cos\left(\dfrac{5\pi}{3}\right) \Bigg)\\\\ &=& 7\Bigg(\cos(5x) + 2\Big( \cos(5x)\cos(\pi)-\sin(5x)\sin(\pi) \Big) \cos\left(\dfrac{5\pi}{3}\right)\Bigg) \\\\ &=& 7\Bigg(\cos(5x) - 2\cos(5x) \cos\left(\dfrac{5\pi}{3}\right) \Bigg)\\\\ &=& 7\Bigg(\cos(5x) \Big(1 - 2\cos\left(\dfrac{5\pi}{3}\right)\Big) \quad | \quad \cos\left(\dfrac{5\pi}{3}\right)=\dfrac{1}{2} \Bigg)\\\\ &=& 7\Bigg(\cos(5x) (1 - 2*\dfrac{1}{2})\Bigg) \\\\ &=& 7\Bigg(\cos(5x) (1 - 1)\Bigg) \\\\ &=& \mathbf{ 0 } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && \mathbf{35\Bigg( \cos(x) + \cos\left(x+\dfrac{2\pi}{3}\right) + \cos\left(x+\dfrac{4\pi}{3}\right) \Bigg)} \\\\ &=& 35\Bigg(\cos(x) + 2\cos\left(\dfrac{2x+2\pi}{2}\right) \cos\left(\dfrac{\dfrac{2\pi}{3}}{2}\right) \Bigg)\\\\ &=& 35\Bigg(\cos(x) + 2\cos(x+\pi) \cos\left(\dfrac{1\pi}{3}\right) \Bigg)\\\\ &=& 35\Bigg(\cos(x) + 2\Big( \cos(x)\cos(\pi)-\sin(x)\sin(\pi) \Big) \cos\left(\dfrac{1\pi}{3}\right)\Bigg) \\\\ &=& 35\Bigg(\cos(x) - 2\cos(x) \cos\left(\dfrac{1\pi}{3}\right) \Bigg)\\\\ &=& 35\Bigg(\cos(x) \Big(1 - 2\cos\left(\dfrac{1\pi}{3}\right)\Big) \quad | \quad \cos\left(\dfrac{1\pi}{3}\right)=\dfrac{1}{2} \Bigg)\\\\ &=& 35\Bigg(\cos(x) (1 - 2*\dfrac{1}{2})\Bigg) \\\\ &=& 35\Bigg(\cos(x) (1 - 1)\Bigg) \\\\ &=& \mathbf{ 0 } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && \mathbf{\cos^7(x) + \cos^7\left(x+\dfrac{2\pi}{3}\right) + \cos^7\left(x+\dfrac{4\pi}{3}\right)} \\ &=& \frac {1}{128} \Bigg( 61\cos(3x) + 0+0+0 \Bigg) \\ &=& \mathbf{\frac {61}{128}*\cos(3x)} \\ \hline \end{array}\)

\(\begin{array}{|rclcrcl|} \hline \mathbf{\cos^7(x) + \cos^7\left(x+\dfrac{2\pi}{3}\right) + \cos^7\left(x+\dfrac{4\pi}{3}\right)} &=& \mathbf{0} \\\\ \frac {61}{128}*\cos(3x) &=& 0 \\\\ \cos(3x) &=& 0 \\ 3x &=& \pm\arccos(0) +2\pi n \quad n \in \mathbb{Z} \\\\ 3x &=& \pm\dfrac{\pi}{2} +2\pi n \\\\ \mathbf{x} &=& \mathbf{\pm\dfrac{\pi}{6} +\dfrac{2\pi n}{3} } \\ \hline \begin{array}{|rclcrcl|} \hline x&=& \dfrac{\pi}{6} +\dfrac{2\pi n}{3} &\text{or}& x&=& -\dfrac{\pi}{6} +\dfrac{2\pi n}{3} \\\\ \mathbf{x_1}&=& \mathbf{\dfrac{\pi}{6}} & & x_4&=& -\dfrac{\pi}{6} +\dfrac{2\pi}{3} \\\\ & & & & \mathbf{x_4}&=& \mathbf{\dfrac{\pi}{2}} \\\\ x_2&=& \dfrac{\pi}{6} +\dfrac{2\pi}{3} & & x_5&=& -\dfrac{\pi}{6} +\dfrac{4\pi}{3} \\\\ \mathbf{x_2}&=& \mathbf{\dfrac{5\pi}{6}} & & \mathbf{x_5}&=& \mathbf{\dfrac{7\pi}{6}} \\\\ x_3&=& \dfrac{\pi}{6} +\dfrac{4\pi}{3} & & x_6&=& -\dfrac{\pi}{6} +\dfrac{6\pi}{3} \\\\ \mathbf{x_3}&=& \mathbf{\dfrac{3\pi}{2}} & & \mathbf{x_6}&=& \mathbf{\dfrac{11\pi}{6}} \\ \hline \end{array} \\ \hline \end{array}\)

Solutions \(0\leq x\leq 2\pi\):

\(\begin{array}{rcll} x &= \dfrac{\pi}{6} \\\\ x &= \dfrac{\pi}{2} \\\\ x &= \dfrac{5\pi}{6} \\\\ x &= \dfrac{7\pi}{6} \\\\ x &= \dfrac{3\pi}{2} \\\\ x &= \dfrac{11\pi}{6} \\ \end{array}\)

Two circles are drawn, with the same center. A chord of the large circle is drawn, so that it is tangent to the small circle.

If the chord has a length of 12 then find the area of the ring-shaped region that is inside the large circle but outside the small circle.

\(\begin{array}{|rcll|} \hline \mathbf{R^2} &=& \mathbf{6^2 + r^2} \\ \mathbf{R^2 - r^2} &=& \mathbf{36} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{\text{area of the ring}} &=& \mathbf{\pi R^2 - \pi r^2} \\ \text{area of the ring} &=& \pi (R^2 - r^2) \quad | \quad \mathbf{R^2 - r^2=36} \\ \mathbf{\text{area of the ring}} &=& \mathbf{36\pi} \\ \hline \end{array} \)

Terry, a marathon runner ran from her house to the high school then back.

The return trip took 5 minutes longer. If her speed was 10 mph to the high school and 9 mph to her house,

How far is Terry's house from the high school?

\(\begin{array}{|rcl|c|rcl|} \hline &(1)& && &(2) \\ \hline \mathbf{v_1} &=& \mathbf{\dfrac{s}{t}} && \mathbf{v_2} &=& \mathbf{\dfrac{s}{t+\dfrac{5}{60} }} \\ && & \boxed{v_1 = 10,\ v_2 = 9 }\\ 10 &=& \dfrac{s}{t} && 9 &=& \dfrac{s}{t+\dfrac{5}{60} } \\\\ \mathbf{t} &=& \mathbf{\dfrac{s}{10}} && 9 &=& \dfrac{s}{t+\dfrac{5}{60} } \\\\ & & && 9\Big(t+\dfrac{5}{60}\Big) &=& s \\\\ & & && 9t+\dfrac{9*5}{60} &=& s \\\\ & & && 9t+0.75 &=& s & \mathbf{t=\dfrac{s}{10}} \\\\ & & && \dfrac{9s}{10}+0.75 &=& s & | \quad * 10 \\\\ & & && 9s + 7.5 &=& 10s \\ & & && \mathbf{s} &=& \mathbf{7.5} \\ \hline \end{array}\)

Terry's house from the high school is 7.5 miles far away.

At 2:00 PM, a car leaves town A and another leaves town B.
The distance between the two towns is 60 km. One car travels 10 kph faster than the other.
The car pass each other at 2:30 PM. How far does each travel?

\(\begin{array}{|rcl|c|rcl|} \hline &(1)& && &(2) \\ \hline v_1 &=& \dfrac{s_1}{t} && v_2 &=& \dfrac{s_2}{t} \\ && & \boxed{v_1 = v_2+10,\\ s_1+s_2=60,\ s_2 = 60-s_1 }\\ v_2+10 &=& \dfrac{s_1}{t} && v_2 &=& \dfrac{60-s_1}{t} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (1)-(2): & v_2+10-v_2 &=& \dfrac{s_1}{t}-\dfrac{60-s_1}{t} \\\\ & 10 &=& \dfrac{s_1}{t}-\dfrac{60-s_1}{t} \quad | \quad *t \\\\ & 10t &=& s_1 - (60-s_1) \\ & 10t &=& 2s_1 - 60 \\ & 2s_1 &=& 10t+60 \quad | \quad :2 \\ & \mathbf{s_1} &=& \mathbf{5t+30} \\ \hline & s_2 &=& 60-s_1 \\ & s_2 &=& 60-(5t+30) \\ & \mathbf{s_2} &=& \mathbf{30-5t} \\ \hline \end{array}\)

\(\text{Let $t=2:30\ PM-2:00\ PM=0.5$ hours}\)

\(\begin{array}{|rcll|} \hline \mathbf{s_1} &=& \mathbf{5t+30} \quad | \quad t=0.5 \\ s_1 &=& 5*0.5+30 \\ \mathbf{s_1} &=& \mathbf{32.5\ \text{km} } \\ \hline \end{array} \begin{array}{|rcll|} \hline \mathbf{s_2} &=& \mathbf{30-5t} \quad | \quad t=0.5 \\ s_2 &=& 30-5*0.5 \\ \mathbf{s_2} &=& \mathbf{27.5\ \text{km} } \\ \hline \end{array}\)

How far does each travel:
One car travel \(\mathbf{32.5\ \text{km}}\) and the other car travel \(\mathbf{27.5\ \text{km}}\)

Simplify

\( \dfrac{\dbinom{n}{k}}{\dbinom{n}{k - 1}}\)

\(\begin{array}{|rcll|} \hline \mathbf{\dbinom{n}{k - 1}} &=& \dfrac{n!}{(k-1)!(n-k+1)!} \quad &| \quad (k-1)!k=k!~ \text{or}~(k-1)!= \dfrac{k!}{k} \\\\ &=& k *\dfrac{n!}{k!(n-k+1)!} \quad &| \quad (n-k)!(n-k+1)=(n-k+1)! \\\\ &=& \dfrac{k}{n-k+1} * \dfrac{n!}{k!(n-k)!} \quad &| \quad \dfrac{n!}{k!(n-k)!} = \dbinom{n}{k} \\\\ &=& \mathbf{ \dfrac{k}{n-k+1} \dbinom{n}{k} } \\ \hline \end{array}\)

\\(\begin{array}{|rcll|} \hline && \mathbf{\dfrac{\dbinom{n}{k}}{\dbinom{n}{k - 1}}} \\\\ &=& \dfrac{\dbinom{n}{k}}{\dfrac{k}{n-k+1} \dbinom{n}{k} } \\\\ &=& \dfrac{n-k+1}{k} * \dfrac{\dbinom{n}{k}}{\dbinom{n}{k} } \\\\ &=& \mathbf{\dfrac{n-k+1}{k}} \\ \hline \end{array}\)

In the figure, each of the three circles is tangent to the other two

and each side of the equilateral triangle is tangent to two of the circles.

If the length of one side of the triangle is 4, what is the radius of each circle?

\(\begin{array}{|rcll|} \hline \mathbf{\tan(30^\circ)} &=& \mathbf{\dfrac{r}{2-r}} \\\\ (2-r)\tan(30^\circ) &=& r \\\\ 2\tan(30^\circ)-r\tan(30^\circ) &=& r \\\\ 2\tan(30^\circ) &=& r+r\tan(30^\circ) \\\\ 2\tan(30^\circ) &=& r\Big(1+\tan(30^\circ)\Big) \\\\ r &=& \dfrac{2\tan(30^\circ)}{1+\tan(30^\circ)} & \boxed{\tan(30^\circ)=\dfrac{ \sqrt{3} } { 3 } }\\\\ r &=& \dfrac{2*\dfrac{ \sqrt{3} } { 3 }}{1+\dfrac{ \sqrt{3} } { 3 }} \\\\ r &=& \dfrac{2\sqrt{3}}{3\Big( 1+\dfrac{ \sqrt{3} } { 3 } \Big) } \\\\ r &=& \dfrac{2\sqrt{3}}{3+ \sqrt{3} } \\\\ r &=& \dfrac{2\sqrt{3}}{\Big(3+ \sqrt{3}\Big) } * \dfrac{ \Big(3- \sqrt{3}\Big) } { \Big(3+ \sqrt{3}\Big) } \\\\ r &=& \dfrac{2\sqrt{3}\Big(3- \sqrt{3}\Big)} { \Big(3+ \sqrt{3}\Big)\Big(3- \sqrt{3}\Big) } \\\\ r &=& \dfrac{2\sqrt{3}\Big(3- \sqrt{3}\Big)} {9-3 } \\\\ r &=& \dfrac{2\sqrt{3}\Big(3- \sqrt{3}\Big)} {6} \\\\ r &=&\dfrac{2}{6}* \sqrt{3}\Big(3- \sqrt{3}\Big) \\\\ r &=&\dfrac{1}{3}* \sqrt{3}\Big(3- \sqrt{3}\Big) \\\\ r &=&\dfrac{1}{3}* \sqrt{3}*3- \dfrac{1}{3}* \sqrt{3}\sqrt{3} \\\\ r &=& \sqrt{3} - \dfrac{1}{3}*3 \\\\ \mathbf{r} &=& \mathbf{\sqrt{3} - 1}& \sqrt{3}=1.73205080757 \\\\ r &=& 1.73205080757 - 1 \\\\ \mathbf{r} &=& \mathbf{0.73205080757} \\ \hline \end{array}\)

The radius of each circle is \(\approx \mathbf{0.732}\)