heureka

Thank you, CPhill

11)

Two circles, centered at A and B are externally tangent to each other, and tangent to a line L. A third circle, centered at C is externally tangent to the first two circles, and the line L. If the radii of circle A and circle B are 9 and 16, respectively, then what is the radius of circle C?

Picture: https://latex.artofproblemsolving.com/4/d/a/4da24a4edb9693476d2766d290f52dd917498a6f.png

\(\text{Pythagorean theorem:} \)

\(\begin{array}{|lrcll|} \hline (1) & (r_a+r_b)^2 &=& (r_b-r_a)^2 + s_1^2 \\ & r_a^2+2r_ar_b+r_b^2 &=& r_b^2-2r_br_a+r_a^2+s_1^2 \\ & 4r_ar_b & = & s_1^2 \\ & \mathbf{s_1} &\mathbf{=}& \mathbf{2\sqrt{r_ar_b}} \\\\ (2) & (r_b+r_c)^2 &=& (r_b-r_c)^2 + s_3^2 \\ & r_b^2+2r_br_c+r_c^2 &=& r_b^2-2r_br_c+r_c^2 + s_3^2 \\ & 4r_br_c &= & s_3^2 \\ & \mathbf{s_3} &\mathbf{=} & \mathbf{2\sqrt{r_br_c}} \\\\ (3) & (r_a+r_c)^2 &=& (r_a-r_c)^2 + s_2^2 \\ & r_a^2+2r_ar_c+r_c^2 &=& r_a^2-2r_ar_c+r_c^2 +s_2^2 \\ & 4r_ar_c & = & s_2^2 \\ & \mathbf{s_2} &\mathbf{=} & \mathbf{2\sqrt{r_ar_c}} \\\\ & \mathbf{s_1} &\mathbf{=}& \mathbf{s_2 + s_3} \\ & 2\sqrt{r_ar_b} &=& 2\sqrt{r_ar_c} + 2\sqrt{r_br_c} \quad & | \quad : 2 \\ & \sqrt{r_ar_b} &=& \sqrt{r_ar_c} + \sqrt{r_br_c} \\ & \sqrt{r_ar_b} &=& \sqrt{r_c}(\sqrt{r_a} + \sqrt{r_b}) \\ & \sqrt{r_c} &=& \dfrac{\sqrt{r_ar_b}} {\sqrt{r_a} + \sqrt{r_b}} \quad & | \quad \Rightarrow \frac{1}{\sqrt{r_c}} = \frac{1}{\sqrt{r_a}} + \frac{1}{\sqrt{r_b}} \\ & r_c &=& \dfrac{ r_ar_b } {(\sqrt{r_a} + \sqrt{r_b})^2} \quad & | \quad r_a = 9 \quad r_b=16 \\ & r_c &=& \dfrac{ 9\cdot 16 } {(\sqrt{9} + \sqrt{16})^2} \\ & r_c &=& \dfrac{ 144 } {(3+4)^2} \\ & r_c &=& \dfrac{ 144 } {49} \\ \hline \end{array}\)

The radius of circle C is \(\mathbf{\tfrac{144}{49}} \)

How many different non-congruent isosceles triangles can be formed by connecting three of the dots in a 4x4 square array of dots like the one shown below?

How many different non-congruent isosceles triangles can be formed by connecting three of the dots in a 4x4 square array of dots like the one shown below?



Two triangles are congruent if they have the same traced outline, possibly up to rotating and flipping. This is equivalent to having the same set of 3 side lengths.

total 9

Wanted to check if essay writer solved my problem X(squared) - 6x - 39 = 0 right:

It is correct!

We proof: \(x_1 = 3-4\sqrt{3}:\)

\(\large{ \begin{array}{|rcll|} \hline x^2-6x-39 &=& 0 \quad & | \quad x_1 = 3 - 4\sqrt{3} \\ (3 - 4\sqrt{3})^2-6(3 - 4\sqrt{3})-39 & \overset{?}{=} & 0 \\ 9 - 24\sqrt{3} + 16\cdot 3 - 18 + 24\sqrt{3} - 39 & \overset{?}{=} & 0 \\ 9 + 16\cdot 3 - 18 - 39 & \overset{?}{=} & 0 \\ 9 + 48 - 18 - 39 & \overset{?}{=} & 0 \\ 57-57 & \overset{?}{=} & 0 \\ 0 & = & 0\ \checkmark \\ \hline \end{array} }\)

We proof \(x_2 = 3+4\sqrt{3}:\)

\(\large{ \begin{array}{|rcll|} \hline x^2-6x-39 &=& 0 \quad & | \quad x_2 = 3 + 4\sqrt{3} \\ (3 + 4\sqrt{3})^2-6(3 + 4\sqrt{3})-39 & \overset{?}{=} & 0 \\ 9 + 24\sqrt{3} + 16\cdot 3 - 18 - 24\sqrt{3} - 39 & \overset{?}{=} & 0 \\ 9 + 16\cdot 3 - 18 - 39 & \overset{?}{=} & 0 \\ 9 + 48 - 18 - 39 & \overset{?}{=} & 0 \\ 57-57 & \overset{?}{=} & 0 \\ 0 & = & 0\ \checkmark \\ \hline \end{array} } \)

A hollow cylinder has radius r cm and height 6r cm, 3 spheres, also of radius r cm, are put into the cylinder

work out the proportion of the cylinder NOT filled by the spheres 

\(\begin{array}{|lcll|} \hline V_{\text{cylinder}} &=& \pi r^2\times 6r \\ &=& 6\pi r^3 \\\\ V_{\text{$3\ $spheres }} &=& 3\times \frac43\pi r^3 \\ &=& 4\pi r^3 \\\\ 1 - \dfrac{4\pi r^3}{6\pi r^3} \\\\ = 1 - \dfrac{2}{3} \\\\ = \dfrac{1}{3} \\ \hline \end{array} \)

\(\text{$\frac13$ of the cylinder is not filled by the spheres.} \)

How do I solve these?

1)

The density of benzene is 0.879 g / mL.

Calculate the mass in grams of 4.25 qt of benzene.

(1 qt = 2 pt, 1 pt = 473 mL)

\(\begin{array}{|rcll|} \hline && 0.879\dfrac{\text{g}}{\text{mL}} \times \dfrac{473\text{ mL}}{1\text{ pt}} \times \dfrac{2\text{ pt}}{1\text{ qt}} \times 4.25\ \text{qt} \\ &=& 0.879\times 473 \times 2 \times 4.25\ \text{g} \\ &=& \mathbf{3534.0195\ \text{g}} \\ \hline \end{array} \)

2)

H is the orthocenter of acute triangle ABC and the extensions of AH, BH, and CH
intersect the circumcircle of traingle ABC at A prime, B prime and C prime.
We know angle AHB : angle BHC : angle CHA = 9 : 10 : 11.
Find angle AprimeBprimeCprime in degrees.
Picture: https://latex.artofproblemsolving.com/6/1/1/6119f02c59ea35f11dceab068d51811cace24c1d.png

\(\text{Let $\angle A'B'C' = \angle CAB = x$ } \\ \text{Let $\angle ACB= \alpha $ } \\ \text{Let $\angle BAC= \beta $ } \\ \text{Let $\angle CBA= \gamma $ } \\ \text{Let $\angle AHB = \gamma + \beta $} \\ \text{Let $\angle BHC = \alpha + \gamma $} \\ \text{Let $\angle CHA = \beta + \alpha $} \\ \text{Let $ \mathbf{x = 180^{\circ}-2 \times\gamma }$} \)

In triangle ABC we set the angles \( \alpha, \beta \text{ and } \gamma\)

further:

\(\mathbf{ \large{x =180^{\circ} - 2 \times \gamma} } \)

\(\begin{array}{|rcrcrcrcrcr|} \hline \angle AHB &:& \angle BHC &:& \angle CHA &=& 9 &:& 10 &:& 11 \\ \gamma + \beta &:& \alpha + \gamma &:& \beta + \alpha &=& 9 &:& 10 &:& 11 \\ \hline \end{array}\)

\(\begin{array}{rclrclrcl} \dfrac{\alpha + \gamma }{\beta + \alpha} &=& \dfrac{10}{11} \qquad & \qquad \dfrac{\gamma + \beta}{\beta + \alpha} &=& \dfrac{9}{11} \\\\ \dfrac{\alpha + \gamma +\gamma + \beta }{\beta + \alpha} &=& \dfrac{10+9}{11} \\\\ \dfrac{\alpha + \beta + 2\times\gamma }{\alpha + \beta } &=& \dfrac{19}{11} \quad &| \quad \alpha + \beta + \gamma = 180^{\circ} \\\\ & & \quad &| \quad \alpha + \beta = 180^{\circ} - \gamma \\\\ \dfrac{180^{\circ} - \gamma + 2\times\gamma }{180^{\circ} - \gamma } &=& \dfrac{19}{11} \\\\ \dfrac{180^{\circ} + \gamma }{180^{\circ} - \gamma } &=& \dfrac{19}{11} \\\\ 11\times(180^{\circ} + \gamma ) &=&19\times(180^{\circ} - \gamma ) \\ 11\times 180^{\circ} + 11\gamma &=&19\times 180^{\circ} - 19\gamma \\ 30\gamma &=&8\times 180^{\circ}\\ \mathbf{\gamma} &\mathbf{=}& \mathbf{48^{\circ}} \\ \end{array} \)

\(\begin{array}{|rcll|} \hline \angle A'B'C' = x &=& 180^{\circ} - 2 \times \gamma \\ \angle A'B'C' &=& 180^{\circ} - 2 \times \gamma \\ &=& 180^{\circ} - 2 \times 48^{\circ} \\ &=& 180^{\circ} - 96^{\circ} \\ &=& \mathbf{84^{\circ}} \\ \hline \end{array} \)

\(\text{The angle $A'B'C'$ in degrees is $\mathbf{84^{\circ}}$ } \)

\(\begin{array}{rcll} \dfrac{\alpha+\gamma}{\alpha+\beta} &=& \dfrac{10}{11} \quad & | \quad \alpha+ \beta=180^{\circ}- \gamma = 180^{\circ}-84^{\circ} = 132^{\circ} \\\\ \dfrac{\alpha+48^{\circ} }{132^{\circ}} &=& \dfrac{10}{11} \\\\ \alpha &=& \dfrac{10\times 132^{\circ}}{11} -48^{\circ} \\\\ \mathbf{ \alpha } & \mathbf{=} & \mathbf{72^{\circ}} \\ \end{array}\)

\(\begin{array}{rcll} \alpha+\beta &=& 132^{\circ} \\ \beta &=& 132^{\circ}-\alpha \\ \beta &=& 132^{\circ}-72^{\circ} \\ \mathbf{ \beta } & \mathbf{=} & \mathbf{60^{\circ}} \\ \end{array}\)

\(\begin{array}{|rcll|} \hline \angle AHB = \gamma + \beta = 108^{\circ} \\ \angle BHC = \alpha + \gamma = 120^{\circ} \\ \angle CHA = \beta + \alpha = 132^{\circ} \\ \hline \end{array} \)

1)
In rectangle ABCD, we have AD = 3 and AB = 4.
Let M be the midpoint of AB, and let X be the point such that MD = MX, angle MDX = 77 degrees,
and A and X lie on opposite sides of DM. Find angle XCD, in degrees.

Picture: https://latex.artofproblemsolving.com/1/b/3/1b39a7de7d5064e27d8ac7d0e00bb8b0db806199.png

\(\text{Let $\angle XCD = x$ } \\ \text{Let $\angle MDX = \angle DXM = 77^{\circ}$ } \\ \text{Let $\angle MDC=\angle MCD = \beta $ } \\ \text{Let $\angle MXC=\angle MCX = \gamma $ } \\ \text{Let $\angle DMC = 180^{\circ} - 2 \times \beta $} \\ \text{Let $\angle XMC = 180^{\circ} - 2 \times \gamma $} \)

\(\begin{array}{|rcll|} \hline \angle DMX &=& 180^{\circ} - 2 \times 77^{\circ} \\ &=& 26^{\circ} \\\\ \angle XMC &=& \angle DMC - 26^{\circ} \\ &=& 180^{\circ} - 2 \times \beta - 26^{\circ} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \angle XMC = 180^{\circ} - 2 \times \beta - 26^{\circ} &=& 180^{\circ} - 2 \times \gamma \\ 180^{\circ} - 2 \times \beta - 26^{\circ} &=& 180^{\circ} - 2 \times \gamma \quad & | \quad \gamma = \beta + x \\ 180^{\circ} - 2 \times \beta - 26^{\circ} &=& 180^{\circ} - 2 \times (\beta + x) \\ 180^{\circ} - 2 \times \beta - 26^{\circ} &=& 180^{\circ} - 2 \times \beta -2\times x \\ - 26^{\circ} &=& -2\times x \\ 26^{\circ} &=& 2\times x \quad & | \quad :2 \\ \mathbf{13^{\circ}} &=& \mathbf{ x } \\ \hline \end{array}\)

The angle XCD is \(\mathbf{13^{\circ}}\)

hallo, please help with this:

\(\large{ \log_x(a)-\log_{\frac1x}(a^3) }\)

\(\begin{array}{|rcll|} \hline && \mathbf{ \large{ \log_x(a)-\log_{\frac1x}(a^3) } } \\ &\large{=}& \large{ \log_x(a)-3\times \log_{\frac1x}(a) } \qquad \boxed{ \log_{\frac1x}(a) = \dfrac{\log_x(a)}{\log_x(\frac{1}{x})} } \\ &\large{=}& \large{ \log_x(a)-3\times \dfrac{\log_x(a)}{\log_x(\frac{1}{x})} } \qquad \boxed{ \log_x\left(\frac{1}{x}\right) = \log_x(1) - \log_x(x) } \\ &\large{=}& \large{ \log_x(a)-3\times \dfrac{\log_x(a)}{ \log_x(1) - \log_x(x) } } \\ &\large{=}& \large{ \log_x(a)-3\times \dfrac{\log_x(a)}{ \log_x(1) - \log_x(x^1) } } \\ && \qquad | \quad \log_x(1) =\log_x(x^0)= 0 \\ && \qquad | \quad \log_x(x^1) = 1 \\ &\large{=}& \large{ \log_x(a)-3\times \dfrac{\log_x(a)}{ 0 - 1 } } \\ &\large{=}& \large{ \log_x(a)+3\times \log_x(a) } \\ &\mathbf{\large{=}}& \mathbf{\large{ 4\times \log_x(a) } }\\ \hline \end{array} \)