ขาหลังข้างซ้ายของแมลงสาบสมัยหิน

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6. The inscribed circle of a triangle $ABC$ touches the sides at $D, E, F,$ and $I$ is its centre. Circles whose radii are $\rho_1,\; \rho_2,\; \rho_3$ are inscribed in the quadrilaterals $IEAF, IFBD, IDCE:$ prove that
$$ \frac{\rho_1}{r-\rho_1} + \frac{\rho_2}{r-\rho_2} + \frac{\rho_3}{r-\rho_3} = \frac{\rho_1 \rho_2 \rho_3}{(r-\rho_1)(r-\rho_2)(r-\rho_3)} \;.$$

Solution:

$\;\;\;$ If the first circle touches $IE$ in $N$, we have $\displaystyle r = IE = IN + NE = \rho_1 \tan \frac{A}{2} + \rho_1,$
therefore $\displaystyle \frac{\rho_1}{r-\rho_1} = \cot \frac{A}{2}.$ But since $\displaystyle \frac{A}{2} + \frac{B}{2} + \frac{C}{2} = \frac{\pi}{2},$ therefore $\displaystyle \Sigma \cot \frac{A}{2} = \Pi \cot \frac{A}{2}.$
Hence the result.

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7. A straight line parallel to $lx + my = 1$ meets the axes (supposed inclined at an angle $\omega$) in $P$ and $Q$. From $P$ and $Q$ perpendiculars are drawn to the lines $py + x = 0,\; qx + y = 0$ respectively. Shew that these perpendiculars meet on a fixed straight line, and find its equation.

Solution:

$\;\;\;$ Let the straight line be $lx + my = k$. Then $P$ is $(\frac{k}{l},0)$ and any line through $P$ is $y = \lambda (x- \frac{k}{l})$. If this is perpendicular to $x + py = 0$, we have

$ \lambda-p = (\lambda p - 1) \cos \omega,\;$ i.e. $\; \lambda (1 - p \cos \omega) = p - \cos \omega.$

Hence the line is

$ (1 - p \cos \omega)y = (p - \cos \omega)(x - \frac{k}{l}).$

Similarly the line through $Q$ is

$ (1 - q \cos \omega)x = (q - \cos \omega)(y - \frac{k}{m}).$

Eliminating $k$, the locus of the intersection is

$ (q - \cos \omega)[(1 - p \cos \omega)l + (p - \cos \omega)m]y = (p - \cos \omega)[(1 - q \cos \omega)m + (q - \cos \omega)l]x $

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มาดูเฉลยกันต่อเลย ...

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8. Two tangents $t_1$ and $t_2$ are drawn to a parabola ; $h$ is the internal bisector of the angle between them and $t$ is the tangent parallel to $h$. Shew that the product of the perpendiculars from the focus on $t_1$ and $t_2$ is the same as that on $t$ and $h$.

Solution:

$\;\;\;$ Let the parabola be $y^2 = 4ax$. Then the equations to $t_1$ and $t_2$ may be taken in the forms
$$ x \cos a_1 + y \sin a_1 + a \sin a_1 \tan a_1 = 0,$$
$$ x \cos a_2 + y \sin a_2 + a \sin a_2 \tan a_2 = 0.$$

The equation to $h$ is found by adding these, and is thus
$$ x \cos \frac{a_1+a_2}{2} + y \sin \frac{a_1+a_2}{2} + a \frac{\sin a_1 \tan a_1 + \sin a_2 \tan a_2}{2 \cos \frac{a_1-a_2}{2}} = 0,$$
whence also the equation to $t$ is
$$ x \cos \frac{a_1+a_2}{2} + y \sin \frac{a_1+a_2}{2} + a \sin \frac{a_1+a_2}{2} \tan \frac{a_1+a_2}{2} = 0.$$

The perpendiculars from $(a, 0)$ on $h$ and $t$ are

$\displaystyle a\left( \cos \frac{a_1+a_2}{2} + \frac{\sin a_1 \tan a_1 + \sin a_2 \tan a_2}{2 \cos \frac{a_1-a_2}{2}} \right)\;\;$ and $\;\; a \cos \frac{a_1+a_2}{2},$

and the product of these is
$$\displaystyle a^2\left( 1+ \frac{\sin a_1 \tan a_1 + \sin a_2 \tan a_2}{ \cos a_1 + \cos a_2} \right) = a^2 \cdot \frac{\sec a_1 + \sec a_2}{\cos a_1 + \cos a_2} = a^2 \sec a_1 \sec a_2,$$
which proves the result, since the perpendiculars from $(a, 0)$ on $t_1$ and $t_2$ are $a \sec a_1$ and $a \sec a_2$.

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9. If the points of intersection of the ellipses $\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1,\; \frac{x^2}{\alpha^2} + \frac{y^2}{\beta^2} = 1$ are at the extremities of conjugate diameters of the former, prove that $\displaystyle \frac{a^2}{\alpha^2} + \frac{a^2}{\beta^2} = 2 \;.$

Solution:

$\;\;\;$ If the points $(a \cos \phi,\; b \sin \phi),\; ( -a \sin \phi,\; b \cos \phi)$ both lie on the second ellipse, we have
$$ \frac{a^2 \cos^2 \phi}{\alpha^2} + \frac{b^2 \sin^2 \phi}{\beta^2} = 1,\;\; \frac{a^2 \sin^2 \phi}{\alpha^2} + \frac{b^2 \cos^2 \phi}{\beta^2} = 1,$$
whence, adding, $\displaystyle \frac{a^2}{\alpha^2} + \frac{a^2}{\beta^2} = 2 \;.$

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10. The centre of gravity of a quadrilateral area is the same as that of four equal particles placed at its angular points : shew that the quadrilateral must be a parallelogram.

Solution:

$\;\;\;$ In general, the centre of gravity of a quadrilateral coincides with that of four equal particles at the angular points and a fifth equal negative particle at the intersection of the diagonals. Hence in this case the centre of gravity of the four equal particles at the angular points must be at the intersection of the diagonals, and this can only be when the diagonals bisect each other, i.e. when the quadrilateral is a parallelogram.

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11. The masses in an Atwood s machine are $5$ lbs. and $3$ lbs., and the machine is being used in a lift which is ascending uniformly with acceleration $2$ ft. per sec. per sec. Find the accelerations of the two masses and the tension of the string.

Solution:

$\;\;\;$ Let $f, f'$ be the downwards and upwards accelerations in space of the masses $5$ lbs. and $3$ lbs., and $T$ the tension of the string. Then $5g -T = 5f,\;\; T-3g = 3f'$.

$\;\;\;$ Also, since the accelerations of the two masses relative to the lift are the same, we have $f+2 = f'-2,\;\;$ i.e. $f'-f = 4.$

Hence $\displaystyle (\frac T 3 - g) - (g - \frac T 5) = 4,\;\;$ i.e. $\displaystyle \frac{8T}{15} = 2g + 4 = 2g$ $+ \frac18g = \frac{17}8g,$

Therefore $T = \frac{255}{64}g,\;\;$ i.e. the tension is $3 \frac{63}{64}$ lbs. wt.

Also, substituting for $T$ in the first two equations, we find $f = \frac{13}2$ and $f' = \frac{21}2$ ft/sec$^2$.

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12. A particle projected with a given velocity from a point on a horizontal plane strikes normally at $P$ a vertical wall at distance a. Shew that if $b$ is the vertical distance between the possible positions of $P$, and $c$ is the height to which the particle would have ascended if projected vertically, then $a^2+b^2 = c^2$.

Solution:

$\;\;\;$ We have $\displaystyle \frac{v^2\sin^2\alpha}{2g} = h,\;\; \frac{v^2\sin\alpha\cos\alpha}{g} = a,\;\; v^2 = 2gc,$
whence the possible values of $h$ are the roots of $\displaystyle \frac{h}{c}\left(1-\frac{h}{c} \right) = \frac{a^2}{4c^2},$ or $\displaystyle h^2 -ch + \frac{a^2}{4} = 0.$

Hence, since $b$ is the difference of the roots, $b^2 = c^2 - a^2$.

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เอาละครับ ... โพสต์เฉลยจบไปอีกหนึ่งชุด

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มาดูเฉลยข้อสอบชุดต่อไปเลยดีกว่า ... ไม่ต้องรอกันนาน

1. If $AA'BB',\; BB'CC',\; CC'AA'$ are three circles, and the straight lines $AA',\; BB',\; CC'$ cut the circle $A'B'C'$ in $\alpha,\; \beta,\; \gamma$ respectively, then the triangle $\alpha \beta \gamma$ will be similar to $ABC$.

Solution:

$\;\;\;$ Since the quadrilaterals $AA'B'B$ and $\alpha A'B' \beta$ ft are both cyclic, therefore
$$A'\hat{A}B = 180^\circ - A'\hat{B'}B = A'\hat{\alpha'}\beta.$$
Therefore $\alpha \beta$ is parallel to $AB$, and similarly for the other sides.

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2. If the diagonals of a quadrilateral circumscribing a conic intersect in a focus, they are at right angles, and the third diagonal is the corresponding directrix.

Solution:

$\;\;\;$ Let $ABCD$ be the quadrilateral, and let each of the tangents from $A$ subtend an angle $\alpha$ at the focus, each of those from $B$ an angle $\beta$, and so on. Then, since $AC$ and $BD$ are straight lines, $\alpha + \beta = \gamma + \delta$. But

$2(\alpha + \beta + \gamma + \delta) = 360^\circ;\;\;$ therefore $\alpha + \beta = 90^\circ,$

i.e. the diagonals are at right angles. Further, since

$2\alpha + 2\beta = 180^\circ,$

each of the lines joining the points of contact of opposite sides is a focal chord. Therefore the extremities of the third diagonal are on the directrix.

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3. If $n$ be any integer, and if the expansion of $(1 + x)^{2n}$ be $1 + a_1x + a_2x^2 + a_3x^3 + ...$, prove that the sum of $n$ terms of the series $a_2 - a_3 + a_4 - a_5 + ...$ is
$$ 2n-1+\frac{(-1)^{n+1}\cdot(2n-1)!}{(n+1)!(n-2)!}.$$

Solution:

$\;\;\;$ The series $1 - a_1 + a_2 - ...$ to $(n + 2)$ terms is the absolute term in the product of $(1 + x)^{2n}$, and the series
$$1 - \frac{1}{x} + \frac{1}{x^2} - ... + (-1)^{n+1}\frac{1}{x^{n+1}} = \frac{1-(-1)^{n+2}\frac{1}{x^{n+2}}}{1+\frac{1}{x}},$$
and is therefore the coefficient of $x^{n+1}$ in $(1+x)^{2n-1}[x^{n+2} - (-1)^{n+2}]$

$\;\;\; = ............................. (-1)^{n+1}(1+x)^{2n-1}$

$\;\;\; = (-1)^{n+1}\frac{(2n-1)!}{(n+1)!(n-2)!}.$

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4. Prove that the problem of dividing a solid hemisphere of unit radius into two equal parts by means of a plane parallel to the base involves the solution of the equation $x^3 - 3x + 1 = 0$, and find graphically the root of this equation which is applicable to the problem in question.

Solution:

$\;\;\;$ Let $x$ be the distance of the cutting plane from the centre. Then, regarding the spherical cap cut off as the difference between a spherical sector and a cone, we have

$\frac13\cdot2\pi(1-x) - \frac13x\pi(1-x^2) = \frac13\pi$,

since the volume of the cap is $\frac14$ that of the sphere.

$\;\;\;$ This equation at once reduces to $x^3 - 3x + 1 = 0$, and the root between $0$and $1$, determined from the graphs $y = x^3$ and $y = 3x-1$, is about $0.35$.

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ถ้าโพสต์โจทย์และเฉลยไปเรื่อยๆ จนจบ (หรือซักครึ่งทางของที่คาดไว้) คิดว่าคงทำลายสถิติหลายอย่างของห้องนี้

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