ขาหลังข้างซ้ายของแมลงสาบสมัยหิน

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10. The faces $AB, AC$ of a fixed triangular wedge make angles $\beta, \gamma$ with the horizon. Equal weights connected by a string which passes without friction over the vertex $A$ rest on the faces, the one weight being just on the point of moving up, the other of moving down. Shew that $\mu = \tan\frac12(\beta-\gamma)$, where $\mu$ is the coefficient of friction for either weight.

Solution:

$\;\;\;$ With the usual notation $\;\;W\sin\beta = T + \mu R = T + \mu W\cos\beta$
and $\;\;T= W\sin\gamma + \mu R' = W\sin\gamma + \mu W\cos\gamma,$
therefore $\;\;W(\sin\beta - \sin\gamma) = \mu W(\cos\beta + \cos\gamma),$
whence $\;\;\mu = \tan \frac12(\beta - \gamma).$

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11. If the radii of a wheel and axle are $1$ ft. and $3$ in., and the suspended weights are $1$ lb., $3$ Ib. respectively, prove that the heavier weight will ascend with acceleration $\frac1{19}g$, the inertia of the machine being neglected.

Solution:

$\;\;\;$ Let $T,\; T'$ be the tensions of the strings round the axle and wheel respectively, and $f, f'$ the accelerations of the weights. Then
$\;\;\;\;\;\; T-3g = 3f,\;\; g-T' =f',\;\; f = \frac14f',\;\; T = 4T',$
whence $\; 3g + 3f = 4g - 4f' = 4g - 16f,$ therefore $f = \frac1{19}g.$

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มาดูเฉลยข้อสุดท้ายของ Papers 3 กันดีกว่า


12. It is required to throw a projectile from a given point below a given plane whose inclination to the horizon is $\alpha$, so as to strike the plane. Shew that the velocity of projection must exceed $(2gc \cos \alpha)^{\frac12}$, where $c$ is the distance of the point from the plane.

Solution:

$\;\;\;$ Taking the point of projection as origin, and $\theta$ as the angle of projection, the equation to the path is
$\;\;\;\;\;\; \displaystyle y = x\tan\theta - \frac12g\cdot \frac{x^2}{V^2 \cos^2 \theta},$
and to the line of the plane $y = x \tan\alpha + c \sec\alpha.$

Eliminating $y,$ we get
$\;\;\;\;\;\; \displaystyle \frac12g\cdot \frac{x^2}{V^2 \cos^2 \theta} + x(\tan\alpha -\tan\theta) + c \sec\alpha = 0.$

Hence, if the intersections are real,
$\;\;\;\;\;\; \displaystyle (\tan\alpha -\tan\theta)^2 > \frac{2g}{V^2 \cos^2 \theta)}\cdot c \sec\alpha,\;\;$ i.e. $\; \displaystyle V^2 > 2cg \cdot \frac{\cos\alpha}{\sin^2 (\alpha-\theta)},$
which is impossible unless $\; \displaystyle V^2 > 2cg \cos\alpha.$

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ว่างๆ จะมาโพสต์โจทย์เพิ่มเติมให้อีก ... ตอนนี้เหลือแค่ Problem Papers 4 เท่านั้นที่ยังไม่เฉลย

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มาดูข้อสอบชุดที่ 5 กันต่อ ... หวังว่าคงไม่ยากจนเกินไป


PROBLEM PAPERS 5

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 $ABC$.

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.

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

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.

5. If $A+ B + C = \pi,\;\; \gamma-\beta = \pi-A,\;\; \beta-\alpha = \pi-C$, prove that
$\;\;\;\;\;\; \sin A \cos 2\alpha + \sin B \cos 2\beta + \sin C \cos 2\gamma + $
$\;\;\;\;\;\;\;\;\;\;\;\; (\sin A + \sin B + \sin C) (\cos (\beta+\gamma) + \cos (\gamma+\alpha) + \cos (\alpha+\beta)) = 0.$

6. On the sides of a triangle as bases are described internally three isosceles triangles with base angles $\theta$. If the triangle formed by their vertices is similar to the original triangle, then
$$\tan\theta = \frac{\sin A + \sin B + \sin C}{1+\cos A \cos B \cos C}.$$

7. From a point $P(x' , y')$ perpendiculars $PM,\; PN$ are drawn on the straight lines $ax^2+2hxy + by^2 = 0$. Shew that if $O$ is the origin, the area of the triangle $OMN$ is
$$\frac{(ay'^2-2hx'y'+bx'^2)(h^2-ab)^\frac12}{(a-b)^2+4h^2}.$$

8. If the normal at $P$ to a parabola meet the curve again in $P$ and the normals at $P,\; P'$ make angles $\alpha,\; \alpha'$ with the axis, prove that $3 \cos\alpha' + \cos (2\alpha-\alpha' ) = 0$.

9. Shew that the four common tangents of the ellipse $\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$ and the rectangular hyperbola $xy = c^2$ are the lines
$$\{4c^2x^2+a^2(2xy-4c^2)\}\{4c^2y^2+b^2(2xy-4c^2)\}+(b^2x^2-a^2y^2)^2 = 0.$$

10. A cylindrical ruler of radius $a$ and length $2l$ rests on a horizontal rail with one end against a smooth vertical wall to which the rail is parallel. Shew that the angle the axis of the ruler makes with the vertical is given by
$$(l \sin\theta + a \cos\theta) \sin^2\theta + 2a \cos\theta = b,$$
where $b$ is the distance of the rail from the wall.

11. An elastic ball is dropped from the ceiling of a room of height $h$, and at the same instant a similar ball is dropped from a point midway between the ceiling and the floor. If the coefficient of restitution in each case is $e$, shew that the balls will first pass each other at a height
$$ \frac{4e^2+4e-1}{8(1+e)^2}h$$
from the floor.

12. A smooth wedge, whose section is an isosceles rightangled triangle, rests on a perfectly smooth table, the hypotenuse being in contact with the table. Two masses $m_1,\; m_2$ are attached to the ends of a string, which passes over a very small pulley at the summit of the wedge. Prove that the wedge moves on the table with acceleration
$$\frac{(m_1-m_2)g}{2M+m_1+m_2},$$
where $M$ is the mass of the wedge.

.
... ขอเชิญยอดฝีมือทุกท่านร่วมถล่มโจทย์ได้เลยครับ

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ติดตามมาด้วยข้อสอบชุดที่ 6 กันเลย


PROBLEM PAPERS 6

1. Two circles are drawn touching the sides $AB,\; AC$ of a triangle $ABC$ at the ends of the base $BC$ and also passing through its middle point $D$. If $E$ be the other point of intersection,
prove that $DA \cdot DE = DC^2$.

2. If a line be drawn through a focus of a central conic, making a constant angle with a tangent, prove that the locus of the point of intersection is a circle.

3. Shew that, if $n$ be a positive integer,
$$(x + y+ z)^{2n+1} - x^{2n+1} - y^{2n+1} - z^{2n+1}$$
is, save for a numerical factor, divisible by
$$(x + y + z)^3 - x^3 - y^3 - z^3.$$

4. Shew that the expression
$$\displaystyle \Sigma_{r=1}^{r=n} \frac{a_r^{m-1}}{(a_r-a_1)(a_r-a_2)...(a_r-a_{r-1})(a_r-a_{r+1})...(a_r-a_n)}$$
is equal to zero if $m< n$, to unity if $m = n$ and to $a_1 + a_2 + ... + a_n$ if $m= n + 1$.

5. By means of a circle and the cosine-graph, or otherwise, find approximately the positive root of the equation $x^2 + \cos^2x = 2$.

6. Prove that $\displaystyle \sin\frac{\pi}{14} \sin\frac{3\pi}{14} \sin\frac{5\pi}{14} = \frac18$.

7. Prove, without assuming anything but the definition, that the straight lines $3x-y = 0$ and $x + 17y = 0$ are conjugate diameters of the conic $2x^2-xy+3y^2 = 1$.

8. The tangent at a point $P$ of a parabola meets another parabola with the same vertex and axis in $Q$ and $R$. Shew that, if $\theta_1,\; \theta_2,\; \theta_3$ are the angles the tangents at $P,\; Q,\; R$ make with the common axis, then $\cot \theta_1$ is the arithmetic mean between $\cot \theta_3$ and $\cot \theta_3$.

9. The tangent at $P$ to the hyperbola $\displaystyle \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$ meets the lines $y^2 = c^2$ at the points $Q$ and $R$. Shew that $c$ can be chosen so that $CP$ bisects the angle $QCR$ for all positions of $P$ on the curve.

10. A weight $W$ is attached to an endless string of length $l$, which hangs over two smooth pegs distant $c$ apart in a horizontal line. Prove that the pressure on each peg is of magnitude
$$W\left(\frac{l-c}{2(l-2c)}\right)^\frac12.$$

11. A piece of wire of given length is bent into the form of a circular quadrant and its two bounding radii : find the position of its centre of gravity.

12. Two heavy particles are projected from a point with equal velocities, their directions of projection being in the same vertical plane: $t, t'$ are the times taken by them to reach the point of intersection of their paths, and $T, T'$ the times to reach their highest point. Shew that $T\;t + T't'$ is independent of the directions of projection.

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ขอบคุณคุณ nongtum ที่ช่วย Message แจ้งแก้โจทย์ข้อ 3 เพราะผมพิมพ์เลขยกกำลังผิดไป ... แก้ให้เรียบร้อยแล้วครับ
ใครเจอข้อสงสัยตรงไหนก็ทักท้วงได้ครับ

.
... ขอเชิญยอดฝีมือทุกท่านร่วมถล่มโจทย์ได้เลยครับ

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ถ้าโพสต์ข้อสอบชุดที่ 7 ต่อทันที ก็ออกจะโหดเกินไป เดี๋ยวเพื่อนๆ จะสำลักโจทย์ซะก่อน
และอีกอย่างคือผมเริ่มเหนื่อยแล้ว

เอาไว้มาสนุกกันต่อครับ

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Question 4 (PROBLEM 5) FIRST HALF

Revolving $ y= \sqrt{1-x^2}$ (in first quadrant) around X-axis

And solve for $a$

$ \pi \int_0^a 1-x^2 \,\, dx = \pi \int_a^1 1-x^2 \,\, dx $ (Equal of volume)

This equation above is equivalent to such cubic equation indicated in the problem.

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ระยะหลัง ผมเริ่มสังเกตว่ามีคนเข้าอ่านกระทู้นี้มากขึ้นกว่าเดิมเยอะ ...
ตอนแรกที่เริ่มตั้งกระทู้นี้ ก็แค่คิดสนุกๆ กับข้อความที่อ่านเจอในหนังสือที่บอกไว้ตอนตั้งกระทู้

ครั้งแรกที่อ่านพบข้อความ "ขาหลังข้างซ้ายของแมลงสาบสมัยหิน" ผมนั่งหัวเราะก๊ากอยู่คนเดียวตั้งนาน
สาเหตุที่ขำมากๆ เพราะผมคิดต่อไปว่า "บริษัทที่ไหนมันจะจ้างคนทำวิจัยเรื่องนี้ไปทำงาน ?"
จะมีสักกี่ครั้งในชีวิตของเขาที่มีคนจ้างให้เป็นผู้เชี่ยวชาญแค่ "ขาหลังข้างซ้ายของแมลงสาบสมัยหิน"

เช้านี้กลับมานั่งคิดเล่นๆ เกี่ยวกับเนื้อหาที่ผมเอามาโพสต์ให้อ่านกัน โจทย์พวกนี้มาจากหนังสือที่พิมพ์ครั้งแรก
เมื่อปี 1904 และฉบับที่ผมเอามาเป็น Second Edition ปี 1914 คิดแล้วก็ตกประมาณ 100 ปีมาแล้ว
เลยนึกเปรียบเทียบกับ "แมลงสาบสมัยหิน" เพราะเป็นการขุดโจทย์ระดับ Fossil กลับมาให้คนรุ่นนี้ได้อ่าน

แต่ผมขุดมาให้อ่านแบบ "เต็มตัว (โจทย์ทุกแนว)" ไม่ใช่แค่ "ขาหลังข้างซ้าย (โจทย์เฉพาะเรื่องใดเรื่องหนึ่ง)"
และยังมีการ "พลิกใต้ท้อง (เฉลยโจทย์ทุกข้อ)" ให้ดูกันเต็มๆ อีกต่างหาก

ฉะนั้นกระทู้นี้คงไม่ถูกครหาว่าเป็น "ขาหลังข้างซ้ายของแมลงสาบสมัยหิน" แน่นอน ...

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มาเริ่มดูเฉลย PROBLEM PAPERS 4 กันต่อดีกว่า


1. $AB, AC$ are two radii of a circle inclined at an angle of $60^\circ$. Upon $AC$ a point $P$ is taken such that a circle can be described with centre $P$ to touch the circle internally and also to touch a circle on $AB$ as diameter externally. Prove that $AP = \frac45AC.$

Solution:

$\;\;\;$ Let $Q$ be the middle point of $AB$, and draw $PN$ per pendicular to $AB$. Call $AB$ unity,
and let $AP = k$. Then
$\;\;\;\;\;\;\;\;\;\;\;\; PQ = \frac12 + (1-k) = \frac32 - k$.
Hence $QN^2 - AN^2 = (\frac32 - k)^2 - k^2 = \frac94 - 3k$
and $QN + AN = \frac12$.
therefore $QN - AN = \frac92 - 6k$, therefore $AN = 3k - 2$.
But $AN = \frac12AP = \frac12k$, therefore $3k-2 = \frac12k$, whence $k = \frac45$.

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2. Given a directrix and two points on a conic, find the locus of the corresponding focus.

Solution:

$\;\;\;$ Let $P, P'$ be the two points, and let $PP'$ meet the directrix in $Z$. Then $SZ$ is the external bisector of $P\hat{S}P'$. Hence, if $PP'$ is divided internally in $Y$ in the ratio in which $Z$ divides it externally, the locus of $S$ is the circle on $YZ$ as diameter.

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3. Prove that if $u = a^2 + ab + b^2$ and $v = a^2 - ab + b^2,$ then

$ 4(a^4+b^4) = 6uv - u^2 - v^2,$
$ 4(a^6+b^6) = 3u^2v + 3uv^2 - u^3 - v^3.$

Solution:

$\;\;\;$ Here $a^2 + b^2 = \frac12(u+v),\;\;\; ab = \frac12(u-v),$

therefore $(a^4+b^4) = (a^2 + b^2)^2 - 2a^2b^2 = \frac14(u+v)^2 - \frac12(u-v)^2.$

Also $(a^6+b^6) = (a^2 + b^2)[(a^2 + b^2)^2-3a^2b^2] = \frac14(u+v)(4uv-u^2-v^2).$

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4. If $x < 1,$ shew that
$$ \frac{x+2}{x^2+x+1} = 2 - x - x^2 + 2x^3 - x^4 - x^5 + ... \;\;.$$

Solution:

$\;\;\;$ Since
$$ \frac{x+2}{x^2+x+1} = \frac{(2+x)(1-x)}{1-x^3} = (2-x-x^2)(1-x^3)^{-1}$$
the result follows from the Binomial Theorem.

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ข้อที่เหลือค่อยมาเฉลยต่อละกัน ... พักเหนื่อยก่อน

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5. Shew that
$$ \cos\frac{2\pi}{15} + \cos\frac{4\pi}{15} + \cos\frac{8\pi}{15} + \cos\frac{14\pi}{15} = \frac12 \;.$$

Solution:

$\;\;\;$ The given expression
$\displaystyle \;\;\;\;\;\; = 2\cos\frac{8\pi}{15}\cos\frac{6\pi}{15} + 2\cos\frac{6\pi}{15}\cos\frac{2\pi}{15}$
$\displaystyle \;\;\;\;\;\; = 2\cos\frac{2\pi}{5}\left(\cos\frac{8\pi}{15} + \cos\frac{2\pi}{15}\right)$
$\displaystyle \;\;\;\;\;\; = 2\cos\frac{2\pi}{5} \cdot 2\cos\frac{\pi}{5}\cos\frac{\pi}{3}$
$\displaystyle \;\;\;\;\;\; = 2\cos\frac{2\pi}{5}\cos\frac{\pi}{5}$
$\displaystyle \;\;\;\;\;\; = 2\sin\frac{\pi}{10}\cos\frac{\pi}{5}$
$\displaystyle \;\;\;\;\;\; = 2 \cdot \frac{\sqrt{5}-1}{4} \cdot \frac{\sqrt{5}+1}{4} = \frac12.$