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Square ABCD has area 12.25.
E is the intersection of BD and the semicircle of diameter AD.
CF is a segment drawn from C and tangnet to the semicircle at F.
Find the area of triangle DEF.

\(\begin{array}{|lll|} \hline\mathbf{Point }~E: \text{ Intersection circle and line }BD \\ \hline \begin{array}{rcll} \mathbf{y_E} &=& \mathbf{-x_E+2r} \\ \mathbf{y_E^2} &=& \mathbf{2rx_E-x_E^2} \\ (-x_E+2r)^2 &=& 2rx_E-x_E^2 \\ x_E^2 - 4rx_E+4r^2 &=& 2rx_E-x_E^2 \\ 2x_E^2-6rx_E+4r^2 &=& 0 \quad | \quad :2 \\ \mathbf{x_E^2-3rx_E+2r^2} &=& \mathbf{0} \\\\ x_E^2 &=& \dfrac{3r\pm \sqrt{9r^2-4*(2r^2)} }{2} \\ x_E^2 &=& \dfrac{3r\pm \sqrt{r^2} }{2} \\ x_E^2 &=& \dfrac{3r\pm r} {2} \\\\ x_E &=& \dfrac{3r-r}{2} \\ x_E &=& \dfrac{2r}{2} \\ \mathbf{x_E} &=& \mathbf{r}\ \checkmark \\\\ x_E &=& \dfrac{3r+r}{2} \\ x_E &=& \dfrac{4r}{2} \\ \mathbf{x_E} &=& \mathbf{2r} \quad | \quad \text{Pont }E ~\neq~ \text{Pont }D \\\\ y_E &=& -x_E+2r \quad | \quad x_E= r\\ y_E &=& -r+2r \\ \mathbf{y_E} &=& \mathbf{r}\ \checkmark \\\\ \mathbf{E} &=& \mathbf{(r,r)} \\ \end{array}\\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{MH}: \\ \hline r^2 &=& MH*CM \quad | \quad CM=r\sqrt{5} \\ r^2 &=& MH*r\sqrt{5} \\ r &=& MH\sqrt{5} \\ \mathbf{MH} &=& \mathbf{\dfrac{r}{\sqrt{5}} } \\ \hline \end{array}\\ \begin{array}{|rcll|} \hline \mathbf{FH}: \\ \hline FH^2 &=& MH*(CM-MH) \\ && \boxed{MH=\dfrac{r}{\sqrt{5}},\ CM=r\sqrt{5}} \\ FH^2 &=& \dfrac{r}{\sqrt{5}}*\left(r\sqrt{5}-\dfrac{r}{\sqrt{5}}\right) \\\\ FH^2 &=& r^2-\dfrac{r^2}{5} \\\\ FH^2 &=& \dfrac{4r^2}{5} \\\\ \mathbf{FH} &=& \mathbf{\dfrac{2r}{\sqrt{5}}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{x_F}: \\ \hline FD^2 &=& (AD-x_F)*AD \quad | \quad FD=2FH,\ AD=2r \\ (2FH)^2 &=& (2r-x_F)*2r \quad | \quad FH =\dfrac{2r}{\sqrt{5}} \\ \left(\dfrac{4r}{\sqrt{5}}\right)^2 &=& (2r-x_F)*2r \\ \dfrac{16r^2}{5} &=& 4r^2-2rx_F \\ 2rx_F &=& 4r^2 - \dfrac{16r^2}{5} \quad | \quad :2r \\ x_F &=& 2r - \dfrac{8r}{5} \\ \mathbf{x_F} &=& \mathbf{\dfrac{2r}{5}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{y_F}: \\ \hline y_F^2 &=& x_F(AD-x_F) \quad | \quad x_F=\dfrac{2r}{5},\ AD=2r \\\\ y_F^2 &=& \dfrac{2r}{5}(2r-\dfrac{2r}{5}) \\\\ y_F^2 &=& \dfrac{4r^2}{5} - \dfrac{4r^2}{25} \\\\ y_F^2 &=& \dfrac{20r^2}{5} - \dfrac{4r^2}{25} \\\\ y_F^2 &=& \dfrac{16r^2}{25} \\\\ \mathbf{y_F} &=& \mathbf{\dfrac{4r}{5}} \\ \hline \end{array}\)

\(\mathbf{F} = \mathbf{(\dfrac{2r}{5},\dfrac{4r}{5})} \)

\(\begin{array}{|lll|} \hline \mathbf{\text{The area of triangle }~DEF}: \\ \begin{array}{lccc} \hline \text{Point} & x & y & \text{cross product} \\ \hline D: & 2r & 0 \\ E: & r & r & 2r*r-r*0 \\ F: & \dfrac{2r}{5} & \dfrac{4r}{5} & r *\dfrac{4r}{5} - \dfrac{2r}{5} * r \\ D: & 2r & 0 & \dfrac{2r}{5} * 0 - 2r * \dfrac{4r}{5} \\ \hline & & & \text{sum }~=2r^2 + \dfrac{4r^2}{5} - \dfrac{2r^2}{5} - \dfrac{8r^2}{5} \\ & & & =2r^2\left(1 + \dfrac{2}{5} - \dfrac{1}{5} - \dfrac{4r}{5}\right) \\ & & & =2r^2\left(1 + \dfrac{2}{5} - 1\right) \\ & & & =2r^2\left( \dfrac{2}{5}\right) \\ & & & =\dfrac{4r^2}{5} \\ \hline \end{array}\\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \text{The area of triangle }~DEF &=& \dfrac{\text{sum}}{2} \\\\ &=& \dfrac{\dfrac{4r^2}{5}}{2} \\\\ &=& \dfrac{2r^2}{5} \\\\ && \text{Square }~ ABCD &= AB*AD \quad | \quad AB=AD=2r \\ && \text{Square }~ ABCD &= (2r)^2 \\ && \text{Square }~ ABCD &= 4r^2 \quad | \quad \text{Square }~ ABCD= 12.25 \\ && 12.25 &= 4r^2 \quad | \quad :4 \\ && 3.0625 &= r^2 \\ && \mathbf{r^2} &= \mathbf{3.0625} \\\\ &=& \dfrac{2*3.0625}{5} \\\\ \mathbf{\text{The area of triangle }~DEF} &=& \mathbf{1.225} \\ \hline \end{array}\)

Square ABCD has area 12.25.

E is the intersection of BD and the semicircle of diameter AD.

CF is a segment drawn from C and tangnet to the semicircle at F.

Find the area of triangle DEF.

\(\begin{array}{|lll|} \hline\mathbf{Point }~E: \text{ Intersection circle and line }BD \\ \hline \begin{array}{rcll} \mathbf{y_E} &=& \mathbf{-x_E+2r} \\ \mathbf{y_E^2} &=& \mathbf{2rx_E-x_E^2} \\ (-x_E+2r)^2 &=& 2rx_E-x_E^2 \\ x_E^2 - 4rx_E+4r^2 &=& 2rx_E-x_E^2 \\ 2x_E^2-6rx_E+4r^2 &=& 0 \quad | \quad :2 \\ \mathbf{x_E^2-3rx_E+2r^2} &=& \mathbf{0} \\\\ x_E^2 &=& \dfrac{3r\pm \sqrt{9r^2-4*(2r^2)} }{2} \\ x_E^2 &=& \dfrac{3r\pm \sqrt{r^2} }{2} \\ x_E^2 &=& \dfrac{3r\pm r} {2} \\\\ x_E &=& \dfrac{3r-r}{2} \\ x_E &=& \dfrac{2r}{2} \\ \mathbf{x_E} &=& \mathbf{r}\ \checkmark \\\\ x_E &=& \dfrac{3r+r}{2} \\ x_E &=& \dfrac{4r}{2} \\ \mathbf{x_E} &=& \mathbf{2r} \quad | \quad \text{Pont }E ~\neq~ \text{Pont }D \\\\ y_E &=& -x_E+2r \quad | \quad x_E= r\\ y_E &=& -r+2r \\ \mathbf{y_E} &=& \mathbf{r}\ \checkmark \\\\ \mathbf{E} &=& \mathbf{(r,r)} \\ \end{array}\\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{MH}: \\ \hline r^2 &=& MH*CM \quad | \quad CM=r\sqrt{5} \\ r^2 &=& MH*r\sqrt{5} \\ r &=& MH\sqrt{5} \\ \mathbf{MH} &=& \mathbf{\dfrac{r}{\sqrt{5}} } \\ \hline \end{array} \begin{array}{|rcll|} \hline \mathbf{FH}: \\ \hline FH^2 &=& MH*(CM-MH) \\ && \boxed{MH=\dfrac{r}{\sqrt{5}},\ CM=r\sqrt{5}} \\ FH^2 &=& \dfrac{r}{\sqrt{5}}*\left(r\sqrt{5}-\dfrac{r}{\sqrt{5}}\right) \\\\ FH^2 &=& r^2-\dfrac{r^2}{5} \\\\ FH^2 &=& \dfrac{4r^2}{5} \\\\ \mathbf{FH} &=& \mathbf{\dfrac{2r}{\sqrt{5}}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{x_F}: \\ \hline FD^2 &=& (AD-x_F)*AD \quad | \quad FD=2FH,\ AD=2r \\ (2FH)^2 &=& (2r-x_F)*2r \quad | \quad FH =\dfrac{2r}{\sqrt{5}} \\ \left(\dfrac{4r}{\sqrt{5}}\right)^2 &=& (2r-x_F)*2r \\ \dfrac{16r^2}{5} &=& 4r^2-2rx_F \\ 2rx_F &=& 4r^2 - \dfrac{16r^2}{5} \quad | \quad :2r \\ x_F &=& 2r - \dfrac{8r}{5} \\ \mathbf{x_F} &=& \mathbf{\dfrac{2r}{5}} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \mathbf{y_F}: \\ \hline y_F^2 &=& x_F(AD-x_F) \quad | \quad x_F=\dfrac{2r}{5},\ AD=2r \\\\ y_F^2 &=& \dfrac{2r}{5}(2r-\dfrac{2r}{5}) \\\\ y_F^2 &=& \dfrac{4r^2}{5} - \dfrac{4r^2}{25} \\\\ y_F^2 &=& \dfrac{20r^2}{5} - \dfrac{4r^2}{25} \\\\ y_F^2 &=& \dfrac{16r^2}{25} \\\\ \mathbf{y_F} &=& \mathbf{\dfrac{4r}{5}} \\ \hline \end{array}\)

\(\mathbf{F} = \mathbf{(\dfrac{2r}{5},\dfrac{4r}{5})} \)

\(\begin{array}{|lll|} \hline \mathbf{\text{The area of triangle }~DEF}: \\ \begin{array}{lccc} \hline \text{Point} & x & y & \text{cross product} \\ \hline D: & 2r & 0 \\ E: & r & r & 2r*r-r*0 \\ F: & \dfrac{2r}{5} & \dfrac{4r}{5} & r *\dfrac{4r}{5} - \dfrac{2r}{5} * r \\ D: & 2r & 0 & \dfrac{2r}{5} * 0 - 2r * \dfrac{4r}{5} \\ \hline & & & \text{sum }~=2r^2 + \dfrac{4r^2}{5} - \dfrac{2r^2}{5} - \dfrac{8r^2}{5} \\ & & & =2r^2\left(1 + \dfrac{2}{5} - \dfrac{1}{5} - \dfrac{4r}{5}\right) \\ & & & =2r^2\left(1 + \dfrac{2}{5} - 1\right) \\ & & & =2r^2\left( \dfrac{2}{5}\right) \\ & & & =\dfrac{4r^2}{5} \\ \hline \end{array}\\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \text{The area of triangle }~DEF &=& \dfrac{\text{sum}}{2} \\\\ &=& \dfrac{\dfrac{4r^2}{5}}{2} \\\\ &=& \dfrac{2r^2}{5} \\\\ && \text{Square }~ ABCD &= AB*AD \quad | \quad AB=AD=2r \\ && \text{Square }~ ABCD &= (2r)^2 \\ && \text{Square }~ ABCD &= 4r^2 \quad | \quad \text{Square }~ ABCD= 12.25 \\ && 12.25 &= 4r^2 \quad | \quad :4 \\ && 3.0625 &= r^2 \\ && \mathbf{r^2} &= \mathbf{3.0625} \\\\ &=& \dfrac{2*3.0625}{5} \\\\ \mathbf{\text{The area of triangle }~DEF} &=& \mathbf{1.225} \\ \hline \end{array}\)

Let \(A_1 A_2 \ldots A_{16}\) be a regular 16-gon.  
Find angle \(A_7 A_9 A_{13}\).

\(\begin{array}{|lrcll|} \hline & 2\beta+ 4*\dfrac{360^\circ}{16} &=& 180^\circ \qquad (1) \\\\ & 2\alpha+ 2*\dfrac{360^\circ}{16} &=& 180^\circ \qquad (2) \\\\ \hline \\ (1)+(2): & 2\beta+ 4*\dfrac{360^\circ}{16} + 2\alpha+ 2*\dfrac{360^\circ}{16} &=& 180^\circ + 180^\circ \\\\ & 2\beta+ 2\alpha + 6*\dfrac{360^\circ}{16} &=& 360^\circ \\\\ & 2\beta+ 2\alpha + 135^\circ &=& 360^\circ \\\\ & 2\beta+ 2\alpha &=& 360^\circ - 135^\circ \\\\ & 2\beta+ 2\alpha &=& 225^\circ \\\\ & 2(\beta+ \alpha) &=& 225^\circ \\\\ & \beta+ \alpha &=& \dfrac{225^\circ}{2} \\\\ & \mathbf{ \beta+ \alpha} &=& \mathbf{112.5^\circ} \\\\ & \theta &=& \beta+ \alpha \\ & \mathbf{ \theta} &=& \mathbf{112.5^\circ} \\ \hline \end{array}\)

\(\angle A_7 A_9 A_{13} = \mathbf{112.5^\circ}\)

Two parallel chords in a circle have lengths 6 and 8. 

The distance between them is 1. 

Find the diameter of the circle.

\(\begin{array}{|rcll|} \hline 3^2+(1+x)^2 &=& R^2 \\ 4^2+x^2&=& R^2 \\ \hline R^2 = 3^2+(1+x)^2 &=& 4^2+x^2 \\ 3^2+(1+x)^2 &=& 4^2+x^2 \\ 9+1+2x+x^2 &=& 16+x^2 \\ 9+1+2x &=& 16 \\ 10+2x &=& 16 \\ 2x &=& 6 \\ \mathbf{x} &=& \mathbf{3} \\\\ R^2 &=& 4^2 + x^2 \\ R^2 &=& 16+3^2 \\ R^2 &=& 25 \\ \mathbf{R} &=& \mathbf{5} \\ \mathbf{2R} &=& \mathbf{10} \\ \hline \end{array}\)

The diameter of the circle is 10

My attempt:

\(4! - 8 \\ = 24-8 \\ = 16\)

If you have four cards with the numbers 1,2,3 and 4
how many numbers can you make that are bigger than 2200?

\(\begin{array}{|rcll|} \hline 1234 & \\ 1243 & \\ 1324& \\ 1342& \\ 1423& \\ 1432& \\ \hline \end{array} \begin{array}{|rcll|} \hline 2134 & \\ 2143 & \\ 2314& \checkmark \\ 2341& \checkmark \\ 2413& \checkmark \\ 2431& \checkmark \\ \hline \end{array} \begin{array}{|rcll|} \hline 3124& \checkmark \\ 3142& \checkmark \\ 3214 & \checkmark \\ 3241 & \checkmark \\ 3412& \checkmark \\ 3421& \checkmark \\ \hline \end{array} \begin{array}{|rcll|} \hline 4123& \checkmark \\ 4132& \checkmark \\ 4213 & \checkmark \\ 4231 & \checkmark \\ 4312& \checkmark \\ 4321& \checkmark \\ \hline \end{array}\)

You can make 16 numbers that are bigger than 2200

If we list the multiples of 3 and 7 under 10 the results would look like this: 3, 6, 7, 9.
The sum of these multiples is 25.
Can you find the sum of the multiples of 3 and 7 under 1000?

\(\begin{array}{|rcll|} \hline 3,\ 6,\ 9,\ \ldots ,\ 999~ (=3*333) \\ 7,\ 14,\ 21,\ \ldots ,\ 994~ (=7*142) \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \text{sum } &=& \dfrac{3+999}{2}*333 + \dfrac{7+994}{2}*142 \\\\ \text{sum } &=& \dfrac{1002}{2}*333 + \dfrac{1001}{2}*142 \\\\ \text{sum } &=& 166833 + 71071 \\\\ \mathbf{\text{sum }} &=& \mathbf{237904} \\ \hline \end{array}\)

In a geometric progression of real numbers, the sum of the first two terms is 7,

and the sum of the first six term is 91. 

Find the sum of the first four terms.

\(\begin{array}{|rclrcl|} \hline \mathbf{s_4 = a \dfrac{1-r^4}{1-r}} && \mathbf{s_2 = a \dfrac{1-r^2}{1-r}} \\ \hline \\ \dfrac{s_4}{s_2} &=& a \dfrac{1-r^4}{1-r} \above 1pt a \dfrac{1-r^2}{1-r} \\\\ \dfrac{s_4}{s_2} &=& \dfrac{(1-r^4)}{1-r^2} \\\\ \dfrac{s_4}{s_2} &=& \dfrac{(1-r^2)(1+r^2)}{1-r^2} \\\\ \dfrac{s_4}{s_2} &=& 1+r^2 \\\\ s_4 &=& s_2(1+r^2) \quad | \quad s_2 = 7 \\\\ \mathbf{s_4} &=& \mathbf{7(1+r^2)} \qquad (1) \\ \hline \end{array} \begin{array}{|rclrcl|} \hline \mathbf{s_2} &=& \mathbf{a+ ar} \\\\ s_2 &=& a(1+r) \quad | \quad s_2 = 7 \\\\ 7 &=& a(1+r) \\\\ \mathbf{a} &=& \mathbf{\dfrac{7}{1+r} } \qquad (2) \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline s_6 &=& S_2 + ar^2+ar^3+ar^4+ar^5 \quad | \quad s_2 = 7, s_6 = 91 \\ 91 &=& 7 + ar^2+ar^3+ar^4+ar^5 \\ 84 &=& ar^2+ar^3+ar^4+ar^5 \\ 84 &=& ar^2 (1 +r+r^2+r^3) \quad | \quad \mathbf{a=\dfrac{7}{1+r} } \\\\ 84 &=& \dfrac{7r^2(1 +r+r^2+r^3)}{1+r} \quad | \quad 1 +r+r^2+r^3 = \dfrac{1-r^4}{1-r} \\\\ 84 &=& \dfrac{7r^2\dfrac{1-r^4}{1-r}}{1+r} \\\\ 84 &=& \dfrac{7r^2(1-r^4)}{(1-r)(1+r)} \\\\ 84 &=& \dfrac{7r^2(1-r^4)}{(1-r^2)} \\\\ 84 &=& \dfrac{7r^2(1-r^2)(1+r^2)}{(1-r^2)} \\\\ 84 &=& 7r^2(1+r^2) \quad | \quad : 7 \\\\ 12 &=& r^2(1+r^2) \\\\ 12 &=& r^2 + r^4 \\\\ r^4+r^2 -12 &=& 0 \\\\ r^2 &=& \dfrac{-1\pm \sqrt{1-4*(-12)} }{2} \\ r^2 &=& \dfrac{-1\pm \sqrt{49} }{2} \\ r^2 &=& \dfrac{-1\pm 7 }{2} \\ \\ r^2 &=& \dfrac{-1+7 }{2} \\ r^2 &=& \dfrac{6 }{2} \\ r^2 &=& \mathbf{3} \\\\ r^2 &=& \dfrac{-1- 7 }{2} \\ r^2 &=& \dfrac{-8 }{2} \\ r^2 &=& -4 \qquad \text{r is a complex number! No solution} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{s_4} &=& \mathbf{7(1+r^2)} \quad | \quad r^2 = 3 \\ s_4 &=& 7(1+3) \\ s_4 &=& 7*4 \\ \mathbf{s_4} &=& \mathbf{28} \\ \hline \end{array}\)

The sum of the first four terms is 28

Thank you, CPhill !

Given that triangle ABCAis an irregular triangle, AD is the median which bisects BC, and the green, blue and red regions are squares.

If the total area of green region and blue region is 9, find the area of the red region.

\(b^2m+c^2n=a(d^2+mn)\)