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Prove that #(s-a)(s-b)(s-c) <= (1/8)abc# ?

Parth S.
Featured 5 months ago

In order to prove this i am assuming that s is the semiperimeter of the triangle #

$s = \frac{a + b + c}{2}$

ok so first let

$s - a = x$
$s - b = y$
$s - c = z$

on solving for $a , b , c$ we get

$a = y + z$
$b = x + z$
$c = x + y$

Now , $\frac{a b c}{8} = \frac{\left(x + y\right) \left(y + z\right) \left(z + x\right)}{8}$

and we know that
# (x+y)/2 >= sqrt(xy) #
As Arithmetic mean of $x , y$ is greater than their geometric mean

so ,
# ((y+z)(x+y)(z+x))/8 >=( (2 sqrt(x y) ) (2 sqrt(yz) )(2 sqrt(zx)))/8 = xyz #
# and xyz=(s-a)(s-b)(s-c) #

Hence

# (abc)/8 >= (s-a)(s-b)(s-c) #

Help! Rotate the axis to eliminate the xy-term in the equation then write the equation ins standard form. #x^2-4xy+4y^2+5sqrt(5)y+1=0# Help!?

Cesareo R.
Featured 5 months ago

$1 + 5 X + 10 Y + 5 {Y}^{2} = 0$

Explanation:

Rotating the coordinate system around the origin by an angle $\theta$ so that

$x = X \cos \left(\theta\right) - Y \sin \left(\theta\right)$
$y = X \sin \left(\theta\right) + Y \cos \left(\theta\right)$

and substituting we have in the conic now in the coordinates $X , Y$

the coefficient of $X Y$ is $3 \sin \left(2 \theta\right) - 4 \cos \left(2 \theta\right)$ so choosing $\theta$ such that $3 \sin \left(2 \theta\right) - 4 \cos \left(2 \theta\right) = 0$ or $\tan \left(2 \theta\right) = \frac{4}{3}$ or $\theta = \frac{1}{2} \arctan \left(\frac{4}{3}\right)$ we will have

$1 + 5 X + 10 Y + 5 {Y}^{2} = 0$ which is a parabola.

CW
Featured 3 months ago

see explanation.

Explanation:

Given $A B = A C , \mathmr{and} \angle C A B = {20}^{\circ}$,
$\implies \angle A B C = \angle A C B = \frac{180 - 20}{2} = {80}^{\circ}$
Given $\angle B D C = {30}^{\circ} , \implies \angle B C D = {70}^{\circ}$
$\implies \angle A C D = 80 - 70 = {10}^{\circ}$

In $\Delta A D C , \frac{A D}{\sin} 10 = \frac{C D}{\sin} 20$
$\implies A D = C D \cdot \sin \frac{10}{\sin} 20 = 0.50771 \times C D$

In $\Delta B D C , \frac{B C}{\sin} 30 = \frac{C D}{\sin} 80$
$\implies B C = C D \cdot \sin \frac{30}{\sin} 80 = 0.50771 \times C D$

Hence, $A D = B C$

proof of $\sin \frac{10}{\sin} 20 = \sin \frac{30}{\sin} 80$

cross multiplying :
$\sin 10 \sin 80 = \sin 20 \sin 30$
$L H S = \sin 10 \cos \left(90 - 80\right) = \sin 10 \cos 10$
$= \frac{1}{2} \cdot \sin 20$ ....(as $\sin x \cos x = \frac{1}{2} \sin 2 x$)
$R H S = \sin 20 \sin 30 = \sin 20 \cdot \frac{1}{2} = \frac{1}{2} \cdot \sin 20$

$L H S = R H S$ (proved)

A parallelogram has sides with lengths of #24 # and #9 #. If the parallelogram's area is #9 #, what is the length of its longest diagonal?

Alan P.
Featured 2 months ago

Longest diagonal: $\textcolor{g r e e n}{\sqrt{657 + 90 \sqrt{23}}} \approx \textcolor{g r e e n}{33.0}$

Explanation:

Since the area is given as $9$ (sq.units), if we use a side with length $9$ as the base,
the height of the parallelogram must be $\textcolor{b r o w n}{1}$

Taking the extension of the base to a point perpendicularly below the opposite vertex of the parallelogram and denoting the length of this extension as $\textcolor{b l u e}{a}$,
by the Pythagorean Theorem
$\textcolor{w h i t e}{\text{XXX}} \textcolor{b l u e}{a} = \sqrt{{24}^{2} - {1}^{2}} = 5 \sqrt{23}$

Denoting the longest diagonal as $\textcolor{red}{b}$
and re-applying the Pythagorean Theorem using the base plus its extension ($9 + \textcolor{red}{a}$) and the height $\textcolor{b r o w n}{1}$
$\textcolor{w h i t e}{\text{XXX}} \textcolor{red}{b} = \sqrt{{\left(9 + \textcolor{b l u e}{5 \sqrt{23}}\right)}^{2} + {\textcolor{b r o w n}{1}}^{2}} = \sqrt{81 + 90 \sqrt{23} + 575 + 1} = \sqrt{657 + 90 \sqrt{23}}$

We can use a calculator to derive the approximation
$\textcolor{w h i t e}{\text{XXX}} \sqrt{657 + 90 \sqrt{23}} \approx 32.99431522$

A line segment is bisected by a line with the equation # 4 y - 6 x = 8 #. If one end of the line segment is at #( 7 , 3 )#, where is the other end?

Alan P.
Featured 2 months ago

Any point on the line $\textcolor{p u r p \le}{2 y - 3 x = 23}$
or
if the given line $\textcolor{red}{4 y - 6 x = 8}$ is to be a perpendicular bisector then #color(brown)(""(-23/15,46/5))#

Explanation:

Apology:
Even omitting some details, this explanation is quite long.

Consider the vertical line $x = 7$ which passes through the given point $\left(7 , 3\right)$
This vertical line will intersect $\textcolor{red}{4 y - 6 x = 8}$ at $\left(7 , \frac{25}{2}\right)$, a point which is $\frac{25}{2} - 3 = \frac{19}{2}$ above the given point #color(green)(""(7,3))#
A point twice as far above #color(green)(""(7,3))# would be at $\left(7 , 3 + 2 \times \frac{19}{2}\right) = \left(7 , 22\right)$
That is the line segment from #color(green)(""(7,3))# to $\left(7 , 22\right)$ is bisected by #color(red)("4y-6x=8)#

Furthermore, as we can see from similar triangles any line parallel to $\textcolor{red}{4 y - 6 x = 8}$ and through $\left(7 , 22\right)$ gives an infinite collection of points, any one of which would serve as an endpoint with #color(green)(""(7,3))# for a line segment bisected by $\textcolor{red}{4 y - 6 x = 8}$

$\textcolor{red}{4 y - 6 x = 8}$ has a slope of $\frac{3}{2}$
so any line parallel to it must also have a slope of $\frac{3}{2}$
and
if such a line passes through $\left(7 , 22\right)$ then we can write its equation using the slope-point form:
$\textcolor{w h i t e}{\text{XXX}} y - 22 = \frac{3}{2} \left(x - 7\right)$
or simplified as
$\textcolor{w h i t e}{\text{XXX}} \textcolor{p u r p \le}{2 y - 3 x = 23}$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

It is possible, since this question was asked under the heading "Perpendicular Bisectors" that it was intended that $\textcolor{red}{4 y - 6 x = 8}$ should be the perpendicular bisector of a derived line segment.

In this case the perpendicular line to $\textcolor{red}{4 y - 6 x = 8}$ passing through #color(green)(""(7,3))#
would have a slope of $- \frac{2}{3}$ (the negative inverse of $\textcolor{red}{4 y - 6 x = 8}$ and (again, working through the slope-point form)
an equation of $\textcolor{b r o w n}{3 y + 2 x = 23}$

The system of equations:
$\textcolor{p u r p \le}{2 y - 3 x = 23}$
$\textcolor{b r o w n}{3 y + 2 x = 23}$
can be solved for the point of intersection: #color(brown)(""(-23/15,46/5))#

Given that three points #A(0,1),P(x,0),B(5,3)# Find the least possible of PA+PB in surd form . can show step cleary?

CW
Featured 2 weeks ago

$x = \frac{5}{4}$
minimum value of $\left(P A + P B\right) = \sqrt{41}$

Explanation:

See Fig 1.
Reflect $B$ over $\text{X-axis}$,
$\implies B \left(5 , 3\right) \to B ' \left(5 , - 3\right)$
$\implies P B = P B ' \text{( reflection symmetry)}$
As long as $P$ moves along the X-axis, $P B = P B '$

See Fig 2.
When $A , P , \mathmr{and} B '$ lie in a straight line, minimum value of $\left(P A + P B '\right)$can be achieved.
When $A , P \mathmr{and} B '$ are collinear, $A P \mathmr{and} P B '$ have the same slope $m$
$\implies m = \frac{0 - 1}{x - 0} = \frac{- 3 - 0}{5 - x}$
$\implies - 3 x = - 5 + x$
$\implies x = \frac{5}{4}$

Hence, min. value of $\left(P A + P B\right) = P A + P B ' = A B '$
$= \sqrt{{\left(5 - 0\right)}^{2} + {\left(- 3 - 1\right)}^{2}}$
$= \sqrt{25 + 16}$
$= \sqrt{41}$

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