Mathematical Tripos Examination for 1878

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วันนี้ฟังแถลงคดียุบพรรคซะเวียนหัวไปหมด ... พรุ่งนี้จะหาเวลามาโพสต์โจทย์และเฉลยในกระทู้ให้

คืนนี้ต้องพักสมองก่อน

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ทิ้งช่วงไปหลายวัน มัวแต่สนุกกับการเฉลยในกระทู้ "ขาหลังข้างซ้ายของแมลงสาบสมัยหิน"
วันนี้ตั้งใจจะเฉลยกระทู้นี้หลายข้อหน่อย ... คิดว่าหลายคนคงติดตามอ่านอยู่

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เฉลยข้อสอบชุดที่ 1: MONDAY, December 31, 1877. 9:00 to 12:00

11. If two chords of a rectangular hyperbola be at right angles, each of their four extremities is the orthocentre of the triangle formed by the other three.
$\;\;\;$ If four such points and the tangent at one of them be given, find the centre of the hyperbola.

Solution:

$\;\;\;$ Let $A, B, C, P$ be four such points, each being the orthocentre of the triangle formed by the other three, and let the tangent at $P$ be given.

$\;\;\;$ (1) The centre of any rectangular hyperbola passing through $A, B, C$ lies on the nine-point circle of $\triangle ABC$. (Besant's Conies, Art.139.)

$\;\;\;$ (2) Let $T'PT$ be the given tangent. Join $AP$, and produce it to meet $BC$ in $L$. From $L$ draw $LQ$ parallel to $T'PT$ to meet the nine-point circle in $Q$. Join $QP$, and produce it to meet the nine-point circle in $O$; then $O$ is the required centre.

$\;\;\;$ For, let $AP$ be bisected in $V$; then $V$ is on the nine-point circle. Join $VO$. Then $V\hat{O}P = V\hat{L}Q = V\hat{P}T$; therefore (Besant's Conies, Art.136) $VO$ is the diameter conjugate to $AP$; therefore $O$ is the centre of the hyperbola.

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เฉลยข้อสอบชุดที่ 1: MONDAY, December 31, 1877. 9:00 to 12:00

12. The section of a cone by a plane, which is not perpendicular to the axis and does not pass through the vertex, is either an ellipse, a parabola, or an hyperbola.
$\;\;\;$ In a given plane are drawn a series of confocal conies, upon which stand cones with their vertical angles right angles. Shew that the locus of their vertices is given by the intersection of an hyperbola, whose vertices are the foci of the conies, and a circle concentric with the hyperbola and passing through its foci.

Solution:

$\;\;\;$ Let the given plane intersect at right angles the plane of the paper in the line $A'HSA$, and let $S, H$ be the given foci. Let $V$ be the vertex of the cone which cuts the plane in the conic whose foci are $S, H$, and vertices $A, A'$, Then the circle inscribed in the triangle $VAA'$ will touch $AA'$ at $S$ or $H$. Let it touch it at $S$, and let it touch $VA, VA'$ at $L, L'$.

Then $A'V - AV = A'L' - AL = A'S- AS =$ constant. Hence $V$ is a point on the hyperbola whose foci are $A, A'$ and vertices $S, H$.

Also, by hypothesis, $A\hat{V}A'$ is a right angle; therefore $V$ lies on the circle of which $AA'$ is a diameter, and this is a circle concentric with the hyperbola, and passing through its foci.

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เอาละครับ ... เฉลยจบทั้ง 12 ข้อของชุดแรกแล้ว
ที่เหลือค่อยหาเวลามาต่อกันอีกที

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มาต่อกันด้วยเฉลยข้อสอบชุดที่ 2 กันดีกว่า ... เริ่มกันเลยครับ

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เฉลยข้อสอบชุดที่ 2: MONDAY, December 31, 1877. 13:30 to 16:00

1. Obtain the value of $\pi$ to six places of decimals from the series
$$\pi = \frac{14}{5}\left[1 + \frac23\left(\frac{1}{50}\right) + \frac{2\cdot4}{3\cdot5}\left(\frac{1}{50}\right)^2 + \frac{2\cdot4\cdot6}{3\cdot5\cdot7}\left(\frac{1}{50}\right)^3+...\right]$$
$$\;\;\;+ \frac{948}{3125}\left[1 + \frac23\left(\frac{9}{6250}\right) + \frac{2\cdot4}{3\cdot5}\left(\frac{9}{6250}\right)^2 + \frac{2\cdot4\cdot6}{3\cdot5\cdot7}\left(\frac{9}{6250}\right)^3+...\right]$$

$\;\;\;$ If $\displaystyle \frac{p}{n}$ and $\displaystyle \frac{n-p}{n}$ be converted into circulating decimals, find the relation between the figures in their periods, $n$ being supposed to be prime to 10 and $p$ less than $n$.

Solution:

$\;\;\;$ (i) The series are
$$\pi = \frac{28}{10}\left[1 + \frac23\left(\frac{2}{100}\right) + \frac{2\cdot4}{3\cdot5}\left(\frac{2}{100}\right)^2 + \frac{2\cdot4\cdot6}{3\cdot5\cdot7}\left(\frac{2}{100}\right)^3+...\right]$$
$$\;\;\;+ \frac{30336}{100000}\left[1 + \frac23\left(\frac{144}{100000}\right) + \frac{2\cdot4}{3\cdot5}\left(\frac{144}{100000}\right)^2 + \frac{2\cdot4\cdot6}{3\cdot5\cdot7}\left(\frac{144}{100000}\right)^3+...\right]$$

$\;\;\;$ First series $= 2.837 941 092,\;$ Second series $= 0.303 651 562,$ therefore $\; \pi = 3.141 592 654.$

$\;\;\;$ Thus, value of $\pi$ to six places is $3.141 593.$ The true value is $3.141
592 653 589...,$ so that the above calculation is correct to the last figure.

$\;\;\;$ (ii) Let $\displaystyle \frac{p}{n} = 0.\alpha\beta\gamma...\xi \;$ and $\;\displaystyle \frac{n-p}{n} = 0.\alpha'\beta'\gamma'...\xi',$
$\;\;\;$ then $\displaystyle \frac{p}{n} = \frac{\alpha\beta\gamma...\xi}{999...9} \;$ and $\;\displaystyle \frac{n-p}{n} = \frac{\alpha'\beta'\gamma'...\xi'}{999...9},$

$\;\;\;$ and, since the sum of the fractions is equal to unity,
$\;\;\;\;\;\;\;\;\; \alpha\beta\gamma...\xi + \alpha'\beta'\gamma'...\xi' = 999...9$

$\;\;\;$ so that, the number of figures in each period being the same,
$\;\;\;\;\;\;\;\;\; \alpha+\alpha' = \beta+\beta' = ... = \xi + \xi' = 9.$

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เฉลยข้อสอบชุดที่ 2: MONDAY, December 31, 1877. 13:30 to 16:00

2. Prove that, if $\phi(x)$ be a rational and integral function of $x$ which vanishes when $x$ is put equal to $a$, then $x-a$ is a factor of $\phi(x)$
$\;\;\;$ Shew that $b^2(a-b)(c-b)\left[(a-b)^2+(c-b)^2\right]-ab^2c(a^2+c^2)+b^5(a-b+c)$ is a complete cube.

Solution:

$\;\;\;$ By actual development the expression is found to be equal to $\{b(-a+b-c)\}^3.$

เขาเฉลยสั้นแค่นี้จริงๆ ครับ

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เฉลยข้อสอบชุดที่ 2: MONDAY, December 31, 1877. 13:30 to 16:00

3. Explain to what extent the equation $a^m \cdot a^n = a^{m+n}$ admits of being proved.
$\;\;\;$ Obtain the values of $a^0$ and $a^{-\frac12}$.
$\;\;\;$ If $\displaystyle \phi(x) = \frac{a^x-a^{-x}}{a^x+a^{-x}}, \;\; F(x) = \frac{2}{a^x+a^{-x}},$ shew that
$$\phi(x+y) = \frac{\phi(x)+\phi(y)}{1+\phi(x)\phi(y)}, \;\; F(x+y) = \frac{F(x)F(y)}{1+\phi(x)\phi(y)}.$$

Solution:

$\;\;\;$ The equations are at once verified by substituting the values of $\phi(x),\; \phi(y),\; F(x),\; F(y)$ in the right-hand members, and performing some slight reductions.

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ข้อ 4 เริ่มกลับสู่การเฉลยยาว ซึ่งเห็นแล้วขี้เกียจ ... ค่อยกลับมาโพสต์ต่อละกัน

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เฉลยข้อสอบชุดที่ 2: MONDAY, December 31, 1877. 13:30 to 16:00

4. Find the sum of the cubes of the first $n$ natural numbers ; and shew that every cube is the difference of two squares, such that if the cube contains an uneven factor $a^3$, each of the squares is divisible by $a^2$.
$\;\;\;$ Find the sum of the cubes of $n$ consecutive terms of an arithmetical progression ; and shew that it is divisible by the sum of the corresponding $n$ terms of the arithmetical progression.

Solution:

$\;\;\;$ (i) Since $1^3+2^3+...+n^3 = \{\frac12n(n+1)\}^2$ it follows that
$\;\;\;\;\;\;\;\;\;\;\;\; n^3 = \{\frac12n(n+1)\}^2 - \{\frac12n(n-1)\}^2.$
$\;\;\;\;\;\;$ Let $n = ab,$ we have to shew that $\frac12b(n+1)$ and $\frac12b(n-1)$ are integers. This is evident, for (1) suppose $n$ even, then since a is uneven, $b$ is even and $\frac12b$ is an integer ; (2) suppose $n$ uneven, then $\frac12(n \pm 1)$ is an integer.

$\;\;\;$ (ii) $a^3+(a+b)^3+...+\{a+(n-1)b\}^3$
$\;\;\;\;\;\; = na^3 + 3a^2b\cdot \frac12n(n-1) + 3ab^2\cdot \frac16\{(n-1)n(2n-1)\} + b^3\cdot \{\frac12n(n-1)\}^2$
$\;\;\;\;\;\; = \frac12n\{2a+(n-1)b\}\{a^2+(n-1)ab+\frac12n(n-1)b^2\}.$

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เฉลยข้อ 4 ค่อยดูน่าอ่านหน่อย แสดงรายละเอียดให้ติดตามได้มากขึ้น

เฉลยข้อ 2 กับ 3 ผมคิดว่าให้รายละเอียดน้อยไป ถึงแม้คนเก่งจะคิดต่อเองได้ก็ตาม

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เฉลยข้อสอบชุดที่ 2: MONDAY, December 31, 1877. 13:30 to 16:00

5. Solve the equations :
$$ x^2-(2a-b-c)x + a^2 +b^2 +c^2-bc-ca-ab = 0;$$
$$ {x^2+2xy-y^2 = ax+by \brace x^2-2xy-y^2 = bx-ay}.$$
$\;\;\;$ If $x^2 = px+q,$ shew that
$$ x^n = \frac{\alpha^n-\beta^n}{\alpha-\beta}x + q\frac{\alpha^{n-1}-\beta^{n-1}}{\alpha-\beta},$$
where $\;\;\; \alpha+\beta = p, \;\; \alpha \beta = -q.$
$\;\;\;$ If $x^3 = px^2 + qx+r,$ express $x^n$ in the form $Ax^2+Bx+C.$

Solution:

$\;\;\;$ (i) The roots of the equation are $;\; a+b\omega+c\omega^2,\;\; a+b\omega^2+c\omega,$
where $\omega$ is an imaginary cube root of unity.

$\;\;\;$ (ii) Adding and subtracting the equations, we have
$\;\;\;\;\;\;\;\;\;\;\;\; 2(x^2-y^2) = (a+b)x+(b-a)y,\;\; 4xy = (a-b)x+(a+b)y,$
which may be written
$\;\;\;\;\;\;\;\;\;\;\;\; \{x-\frac14(a+b)\}^2 - \{y-\frac14(a-b)\}^2 = \frac14ab,$
$\;\;\;\;\;\;\;\;\;\;\;\; \{x-\frac14(a+b)\} \{y-\frac14(a-b)\} = \frac1{16}(a^2-b^2);$
or, putting $\; x-\frac14(a+b) = u,\; y-\frac14(a-b) = v,\;$ these are $u^2-v^2 = \frac14ab,$
$uv = \frac1{16}(a^2-b^2),\;$ which give on solution
$\displaystyle \;\;\; {u = -\frac14(a+b) \brace v = -\frac14(a-b)},\;\; {u = \frac14(a+b) \brace v = \frac14(a-b)},\;\; {u = \frac14i(a-b) \brace v = -\frac14i(a+b)},\;\; {u = -\frac14i(a-b) \brace v = \frac14(a+b)},$
(where $i^2 = -1,$ as usual), leading to
$\displaystyle \;\;\; {x = 0 \brace y = 0},\;\; {x = \frac12(a+b) \brace y = \frac12(a-b)},\;\; {x = \frac14\{(1+i)a+(1-i)b\} \brace y = \frac14\{(1-i)a-(1+i)b\}},\;\; {x = \frac14\{(1-i)a+(1+i)b\} \brace y = \frac14\{(1+i)a-(1-i)b\}}.$

$\;\;\;$ (iii) Let $x^n = Ax + B,$ then we have
$\;\;\;\;\;\;\;\;\;\;\;\; x^n -Ax + B = 0,\; \alpha^n -A\alpha + B = 0,\; \beta^n -A\beta + B = 0,$
whence, eliminating $A$ and $B,$
$\;\;\;\;\;\;\;\;\;\;\;\; x^n(\alpha-\beta) - x(\alpha^n-\beta^n) + \alpha \beta(\alpha^{n-1}-\beta^{n-1}) = 0,$
which gives at once the required equation.

$\;\;\;$ (iv) The result is obtained by the elimination of $A, B, C$ from the equations
$\;\;\;\;\;\;\;\;\;\;\;\; x^n = Ax^2+Bx+C,\; \alpha^n = A\alpha^2+B\alpha+C,\; \beta^n = A\beta^2+B\beta+C,\; \gamma^n = A\gamma^2+B\gamma+C,$
where $\alpha,\; \beta,\; \gamma$ are the roots of $x^3-px^2-qx-r = 0$. It is of course best expressed as a determinant.

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เฉลยข้อ 5 ยาวมากจริงๆ ยิ่งผมยังพิมพ์ปีกกาใหญ่ไม่คล่องด้วย (เพิ่งหัดใช้) เลยรู้สึกว่าพิมพ์เฉลยยากมาก
แต่ก็เข็นจนสำเร็จได้ หวังว่าเพื่อนๆ คงได้รับประโยชน์คุ้มกับแรงที่ผมลงไป

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เฉลยข้อสอบชุดที่ 2: MONDAY, December 31, 1877. 13:30 to 16:00

6. Prove that the logarithm of the product of any number of quantities is equal to the sum of the logarithms of the quantities.
$\;\;\;$ Given that $a^2 + b^2 = 1$ and that
$\;\;\;\;\;\; \log 2 = 0.3010300, \;\;\log (1 + a) = 0.1928998, \;\;\log (1+b) = 0.2622226,$
$\;\;\;$ find $\log(1 + a + b).$

Solution:

$\;\;\;$ Since $a^2 + b^2 = 1,$ we have, $1 + a + b = \sqrt{2 (1 + a) (1+b)},$
as is evident by squaring both sides of the equation. Thus
$\;\;\; \log (1 + a + b) = \frac12 \cdot \{0.3010300 + 0.1928998 + 0.2622226\} = 0.3780762$.
[The numbers in the question correspond to $a = \sin 34^\circ$.]

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เฉลยข้อสอบชุดที่ 2: MONDAY, December 31, 1877. 13:30 to 16:00

7. Define the Trigonometrical Ratios of an angle so as to be true for all magnitudes of the angle ; and make a table shewing the values of the trigonometrical ratios in terms of any one of them.
$\;\;\;$ Prove that the equation $\tan x = kx$ has an infinite number of real roots.

Solution:

$\;\;\;$ As $x$ increases from $n\pi-\frac12\pi$ to $n\pi+\frac12\pi, \tan x$ increases through all real values from $-\infty $ to $\infty $, and therefore passes through a root of the equation $\tan x = kx$; and to every positive root there is a corresponding negative root.