heureka

Geometry problem

\(\text{Let $RA=\dfrac{2}{5}AC $} \\ \text{Let $CR=\dfrac{3}{5}AC $} \\ \text{Let $AP=\dfrac{1}{4}AB $} \\ \text{Let $AQ=\dfrac{3}{4}AB $} \)

\(\begin{array}{|rcll|} \hline 2[ABC] &=& AB*AC*\sin(A) \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline 2[ACQ] &=& AC*\dfrac{3}{4}AB*\sin(A) \\ 2[ACQ] &=& \dfrac{3}{4}AB*AC*\sin(A) \\ 2[ACQ] &=& \dfrac{3}{4}*2[ABC] \\ \mathbf{2[ACQ]} &=& \mathbf{\dfrac{3}{2}[ABC]} \\ \hline \end{array} \begin{array}{|rcll|} \hline [BQC] &=& [ABC]-[ACQ] \quad | \quad * 2 \\ 2[BQC] &=& 2[ABC]-2[ACQ] \\ 2[BQC] &=& 2[ABC]-2[ACQ] \quad | \quad \mathbf{2[ACQ]=\dfrac{3}{2}[ABC]} \\ 2[BQC] &=& 2[ABC]- \dfrac{3}{2}[ABC] \quad | \quad : 2 \\ [BQC] &=& [ABC]- \dfrac{3}{4}[ABC] \\ \mathbf{[BQC]} &=& \mathbf{\dfrac{1}{4}[ABC]} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline 2[ARQ] &=& \dfrac{2}{5}AC*\dfrac{3}{4}AB*\sin(A) \\ 2[ARQ] &=& \dfrac{3}{10}AB*AC*\sin(A) \\ 2[ARQ] &=& \dfrac{3}{10}*2[ABC] \\ \mathbf{2[ARQ]} &=& \mathbf{\dfrac{3}{5}[ABC]} \\ \hline \end{array} \begin{array}{|rcll|} \hline [CRQ] &=& [ACQ]-[ARQ] \quad | \quad * 2 \\ 2[CRQ] &=& 2[ACQ]-2[ARQ] \\ 2[CRQ] &=& 2[ACQ]-2[ARQ] \\ 2[CRQ] &=& \dfrac{3}{2}[ABC]- \dfrac{3}{5}[ABC] \quad | \quad : 2 \\ [CRQ] &=& \dfrac{3}{4}[ABC]- \dfrac{3}{10}[ABC] \\ [CRQ] &=& \left(\dfrac{3}{4}- \dfrac{3}{10}\right) [ABC] \\ [CRQ] &=& \left(\dfrac{30-12}{40}\right) [ABC] \\ [CRQ] &=& \left(\dfrac{18}{40}\right) [ABC] \\ \mathbf{[CRQ]} &=& \mathbf{\dfrac{9}{20}[ABC]} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \dfrac{[BQC]}{[CRQ]} &=& \dfrac{\dfrac{1}{4}[ABC]}{\dfrac{9}{20}[ABC]} \\\\ \dfrac{[BQC]}{[CRQ]} &=& \dfrac{\dfrac{1}{4}}{\dfrac{9}{20}} \\\\ \dfrac{[BQC]}{[CRQ]} &=& \dfrac{1}{4} *\dfrac{20}{9} \\\\ \mathbf{\dfrac{[BQC]}{[CRQ]}} &=& \mathbf{\dfrac{5}{9} } \\ \hline \end{array}\)

If
\(x(a + b) + y(a - b) = 2\)
and
\(ax + by = \dfrac{a^2 + b^2} {a^2 - b^2}\),
then express \(\dfrac{1}{x} + \dfrac{1}{y}\) in terms of \(a\) and \(b\).

\(\begin{array}{|rcll|} \hline \mathbf{x(a + b) + y(a - b)} &=& \mathbf{2} \\\\ y(a - b) &=& 2-x(a + b) \\\\ \mathbf{y} &=& \mathbf{\dfrac{2-x(a + b)} {a - b}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{ax + by } &=& \mathbf{ \dfrac{a^2 + b^2} {a^2 - b^2}} \\\\ ax + b\left( \dfrac{2-x(a + b)} {a - b} \right) &=& \dfrac{a^2 + b^2} {a^2 - b^2} \\\\ ax + b\left( \dfrac{2-x(a + b)} {a - b} \right) &=& \dfrac{a^2 + b^2} {(a-b)(a+b)} \quad | \quad *(a-b) \\\\ a(a-b)x + b\Big(2-x(a + b) \Big) &=& \dfrac{a^2 + b^2} {a+b} \\\\ a(a-b)x + 2b-xb(a + b) &=& \dfrac{a^2 + b^2} {a+b} \\\\ x\Big( a(a-b)-b(a + b)\Big) + 2b &=& \dfrac{a^2 + b^2} {a+b} \\\\ x( a^2-2ab-b^2) + 2b &=& \dfrac{a^2 + b^2} {a+b} \\\\ x( a^2-2ab-b^2) &=& \dfrac{a^2 + b^2} {a+b} -2b \\\\ x( a^2-2ab-b^2) &=& \dfrac{a^2 + b^2-2b(a+b)} {a+b} \\\\ x( a^2-2ab-b^2) &=& \dfrac{a^2 + b^2-2ab-2b^2} {a+b} \\\\ x( a^2-2ab-b^2) &=& \dfrac{a^2 -2ab -b^2} {a+b} \quad | \quad :a^2-2ab-b^2 \\\\ x&=& \dfrac{1} {a+b} \\\\ \mathbf{\dfrac{1}{x}} &=& \mathbf{a+b} \\ \hline \mathbf{y} &=& \mathbf{\dfrac{2-x(a + b)} {a - b}} \\\\ \dfrac{1}{y} &=& \dfrac {a - b} {2-x(a + b)} \quad | \quad \mathbf{x=\dfrac{1} {a+b}} \\\\ \dfrac{1}{y} &=& \dfrac {a - b} {2-\dfrac{1} {(a+b)}(a + b)} \\\\ \dfrac{1}{y} &=& \dfrac {a - b} {2-1} \\\\ \mathbf{\dfrac{1}{y}} &=& \mathbf{a-b} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \dfrac{1}{x} + \dfrac{1}{y} &=& (a+b) + (a-b) \\\\ \mathbf{\dfrac{1}{x} + \dfrac{1}{y}} &=& \mathbf{2a} \\ \hline \end{array}\)

What is the vector that results from transforming \(\vec{a}\) by the transformation matrix?
\(\vec{a} = \dbinom{3}{-1}\)
\(B = \begin{pmatrix} 7 & 3 \\ -6 & -4 \end{pmatrix}\)

My attempt:

\(\begin{array}{|rcll|} \hline \mathbf{\vec{a}B} &=& \mathbf{\dbinom{3}{-1}\begin{pmatrix} 7 & 3 \\ -6 & -4 \end{pmatrix}} \\\\ \vec{a}B &=& \dbinom{3*7+(-1)*(-6)}{3*3+(-1)(-4)} \\\\ \mathbf{\vec{a}B} &=& \mathbf{\dbinom{27}{13}} \\ \hline \end{array}\)

If f(2x) = 2/(2 + x), then find 2f(x).

Set \(x = \frac{x}{2}\)

\(\begin{array}{|rcll|} \hline f(2*\frac{x}{2}) &=& \dfrac{2}{2 + \frac{x}{2}} \\\\ f(x) &=& \dfrac{2}{\frac{4+x}{2}} \\\\ f(x) &=& \dfrac{4}{4+x} \\ \\ 2f(x) &=& 2*\dfrac{4}{4+x} \\\\ \mathbf{2f(x)} &=& \mathbf{\dfrac{8}{4+x}} \\ \hline \end{array}\)

What is the vector that results from transforming \(\vec{a}\) by the transformation matrix?
\(\vec{a} = \dbinom{5}{-3}\)

\(B = \begin{pmatrix} 7 & -6 \\ 3 & -4 \end{pmatrix}\)

My attempt:

\(\begin{array}{|rcll|} \hline \mathbf{\vec{a}B} &=& \mathbf{\dbinom{5}{-3}\begin{pmatrix} 7 & -6 \\ 3 & -4 \end{pmatrix}} \\\\ \vec{a}B &=& \dbinom{5*7+(-3)*3}{5*(-6)+(-3)*(-4)} \\\\ \mathbf{\vec{a}B} &=& \mathbf{\dbinom{26}{-18}} \\ \hline \end{array}\)

In the diagram, BD=5, CE=4, [ABC]= 24 and [ADE]= 18. Find [ACD].

If

\(\cos(x) = \dfrac{2ab}{a^2 + b^2}\),

then find \(\tan\left(\frac{x}{2}\right)^2\) in terms of a and b.

I assume: \(\tan\left(\frac{x}{2}\right)^2 = \tan^2\left(\frac{x}{2}\right)\)

\(\begin{array}{|rcll|} \hline \mathbf{\tan^2\left(\frac{x}{2}\right)} &=& \mathbf{\dfrac{1-\cos(x)}{1+\cos(x)}} \\\\ \tan^2\left(\frac{x}{2}\right) &=& \dfrac{1-\dfrac{2ab}{a^2 + b^2}}{1+\dfrac{2ab}{a^2 + b^2}} \\\\ \tan^2\left(\frac{x}{2}\right) &=& \dfrac{\dfrac{a^2 + b^2-2ab}{a^2 + b^2}}{\dfrac{a^2 + b^2+2ab}{a^2 + b^2}} \\\\ \tan^2\left(\frac{x}{2}\right) &=& \dfrac{a^2 + b^2-2ab}{a^2 + b^2+2ab} \\\\ \tan^2\left(\frac{x}{2}\right) &=& \dfrac{\left(a-b \right)^2}{\left(a+b \right)^2} \\\\ \mathbf{\tan^2\left(\frac{x}{2}\right)} &=& \mathbf{\left(\dfrac{a-b}{a+b} \right)^2} \\ \hline \end{array}\)

The polar equation r=7sin(2θ) graphs as a rose. What is the length of the petals of this rose?

The graph:

The fifth term of an arithmetic sequence is 9 and the 32nd term is -84.
What is the 23rd term?

\(\begin{array}{|rcll|} \hline i < j < k \\ a_i &=& a+(i-1)d \\ a_j &=& a+(j-1)d \\ a_k &=& a+(k-1)d \\ \hline \mathbf{a_j-a_i } &=& \mathbf{(j-1)d-(i-1)d} \\ a_j-a_i&=& d\Big((j-1)-(i-1)\Big) \\ \mathbf{a_j-a_i} &=& \mathbf{ d(j-i) } \\ \\ \mathbf{a_k-a_i} &=& \mathbf{(k-1)d-(i-1)d} \\ a_k-a_i&=& d\Big((k-1)-(i-1)\Big) \\ \mathbf{a_k-a_i} &=& \mathbf{ d(k-i) } \\ \hline \dfrac{a_j-a_i}{a_k-a_i} &=& \dfrac{ d(j-i)}{d(k-i)} \\\\ \dfrac{a_j-a_i}{a_k-a_i} &=& \dfrac{ (j-i)}{(k-i)} \\ (a_j-a_i)(k-i) &=& (a_k-a_i)(j-i) \\ a_j(k-i)-a_i(k-i) &=& a_k(j-i)-a_i(j-i) \\ a_j(k-i) &=& a_k(j-i)+a_i(k-i)-a_i(j-i) \\ a_j(k-i) &=& a_k(j-i)+a_i\Big(k-i)-(j-i)\Big) \\ a_j(k-i) &=& a_k(j-i)+a_i(k-j) \\ \mathbf{a_j} &=& \mathbf{a_k\frac{j-i}{k-i}+a_i\frac{k-j}{k-i}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{a_j} &=& \mathbf{a_k\frac{j-i}{k-i}+a_i\frac{k-j}{k-i}} \\ && \boxed{a_{5} =9=a_i\quad i = 5\\ a_{23}=a_j \quad j=23\\ a_{32}=-84=a_k\quad k=32} \\\\ a_{23} &=& (-84)*(\frac{23-5}{32-5})+ 9*(\frac{32-23}{32-5}) \\\\ a_{23} &=& (-84)*(\frac{18}{27})+ 9*(\frac{9}{27}) \\\\ a_{23} &=&(-84)*(\frac{2}{3})+ 9*(\frac{1}{3}) \\\\ a_{23} &=& -\frac{84*2}{3}+ \frac{9}{3} \\\\ a_{23} &=& - 28*2 + 3 \\\\ \mathbf{a_{23}} &=& \mathbf{-53} \\ \hline \end{array}\)

The 23rd term is -53

Find the coefficient of \(x^2 y^2 z^3\) in the expansion of \((4x + 5y - 3z)^7\).

Multinomial theorem:

For any positive integer m and any nonnegative integer n,

the multinomial formula tells us how a sum with m terms expands when raised to an arbitrary power n:

\(\left( x_1+x_2+\cdots +x_m \right)^n = \sum \limits_{k_1+k_2+\dots+k_m=n} \dbinom{n}{k_1,k_2,\dots,k_m}\cdot x_1^{k_1}\cdot x_2^{k_2}\dots x_m^{k_m}\)

where

\(\dbinom{n}{k_1,k_2,\dots,k_m} = \dfrac{n!}{k_1!k_2!\dots k_m!} \)

\(\begin{array}{|lcll|} \hline x^2 y^2 z^3 \\ n=7,\ m=3 \\ \hline k_1 = 2,\ k_2 = 2,\ k_3 = 3 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline (4x + 5y - 3z)^7 &=& \dots + \dfrac{7!}{2!2!3!} (4x)^2(5y)^2(-3z)^3 + \dots \\ \hline && \dfrac{7!}{2!2!3!} (4x)^2(5y)^2(-3z)^3 \\\\ &=& \dfrac{7!}{2!2!3!}* 16x^225y^2(-27)z^3 \\\\ &=& \dfrac{4*5*6*7}{4}* 16*25*(-27)x^2y^2z^3 \\\\ &=& -210* 10800*x^2y^2z^3 \\\\ &=& \mathbf{-2268000}x^2y^2z^3 \\\\ \hline \end{array}\)