มาดูกันครับว่า ... แนวคิดในการทำข้อสอบชุดที่ 1 แต่ละข้อเป็นอย่างไรบ้าง
ขอแนะนำน้องๆ ให้ลองคิดเองดูก่อน และใช้แนวคิดนี้ในการตรวจสอบภายหลัง
SOLUTION TO PROBLEM PAPERS 1
Let $P$ be the given point and $A$ the intersection of the given straight lines. Take points $B$ and $C$, one on each line, such that $P$ lies in the angle $BAG$, and $AB = AC =$ half the given perimeter. Describe $A$ circle touching the lines at $B$ and $C$, and from $P$ draw tangents to this circle. Then evidently either of these tangents fulfils the required condition, provided $P$ lies between $A$ and the circle. There are also, in any case, two other solutions, obtained by drawing the inner tangents from $P$ to the similarly described circles in the angles adjacent to that containing $P$.
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Let $S, S?$ be the centres of the two circles, $a$ and $\frac12 a$ their radii, $P$ the centre of the variable circle and $r$ its radius. Then $SP = a-r, S?P=a + r,$ therefore $SP + S?P = \frac32 a$. Hence the locus of $P$ is an ellipse with $S, S?$ as foci, and whose major axis is $\frac32 a$. Also, since $SS? = \frac12 a$, it easily follows that the eccentricity is $\frac13$, a the semi-minor axis $\frac{a}{\sqrt 2}$, and the latus-rectum $\frac43 a$.
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Evidently $b-c, c-a, a-b$ are factors, and the remaining factor must be a symmetrical homogeneous function of $a, b, c$ of the third degree. We may therefore assume that the given expression is equal to
$$\prod (b-c) \cdot \left[ p \cdot \sum a^3 + q \cdot \sum a^2b + r \cdot abc \right],$$
where $p, q, r$ are numerical constants to be determined.
Putting $c = 0,$ we get
$$a^5b-b^5a = b(-a)(a-b) \cdot \left[ p (a^3 + b^3) + q (a^2b + ab^2) \right].$$
But since $a^5b-b^5a = ab(a-b)(a+b)(a^2 + b^2),$ it follows that $p = q= -1$. To find $r,$ put $a = 2, b=1, c= -1$. We then find $4p + 2q-r = -6,$ whence $r = 0$.
Hence the remaining factor is $-(\sum{a^3}+\sum{a^2b}=-(\sum a)(\sum a^2)$.
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We have $\left(\frac34\right)^{\frac45}= \left(1-\frac14\right)^{\frac45}=1+\frac45(-\frac14)+\frac{\frac45(\frac45-1)}{2!} (-\frac14)^2+? = 1-u_1-u_2-u_3-?$ say.
Then
$u_1 = 0.200000$
$u_2 = \frac{1}{10} \cdot \frac14 u_1 = 0.005000$
$u_3 = \frac25 \cdot \frac14 u_1 = 0.000500$
$u_4 = \frac{11}{20} \cdot \frac14 u_1 = 0.000069$
$u_5 = \frac{16}{25} \cdot \frac14 u_1 = 0.000011$
$u_6 = \frac{7}{10} \cdot \frac14 u_1 = 0.000002$
$u_1 + u_2 + u_3 + u_4 + u_5 + u_6 = 0.205582$.
Hence
$\left(\frac34\right)^{\frac45} = 1- 0.205582... = 0.7944$ correct to four decimal places.
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The left-hand side
$= \frac12 (4 - \cos 24^\circ - \cos 42^\circ - \cos 78^\circ - \cos 96^\circ)$
$= \frac12 (4 - 2 \cos 60^\circ \cos 36^\circ - 2 \cos 60^\circ \cos 18^\circ)$
$= \frac12 (4 - \cos 18^\circ - \cos 36^\circ)$
$= 1 + \sin^2 9^\circ + \sin^2 18^\circ.$
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Here $\sum{PA_1} = 2R \cdot \sum_{r=0}^{r=n-1}\sin\left(\alpha+\frac{r\pi}{n}\right)$
$= 2R \frac{\sin\left(\alpha+\frac{(n-1)\pi}{2n}\right)\sin\left(n\cdot\frac{\pi}{2n}\right)}{\sin\frac{\pi}{2n}}$
$= 2R\cdot \frac{\cos\left(\alpha -\frac{\pi}{2n}\right)}{\sin\frac{\pi}{2n}}$
$= 2R\left(\cos\alpha \cot\frac{\pi}{2n}+\sin\alpha\right).$
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The equation to any conic circumscribing this triangle must be of the form
$(y+kx)(y-kx)+(x\cos\alpha+y\sin\alpha-p)(lx+my) = 0 ...(i).$
Using the conditions for a circle, we find
$l\cos\alpha-k^2 = m \sin\alpha+1$ and $l\sin\alpha + m \cos\alpha = 0,$
whence $\frac{l}{\cos\alpha} = \frac{m}{-\sin\alpha} = 1+k^2.$
Substituting in $(i)$ and simplifying, we obtain the equation given.
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If $P$ be the point $m$, and $P?$ the point $m?$, the equations to $PP?$ and $P?Q$ are respectively
$2x-(m+m?)y+2\alpha m m? = 0$ and $2x-(m?-m)y-2\alpha m m? = 0,$
and if the first of these passes through $(-a, 0),$ the second passes through $(a, 0)$.
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Let the true weight of the first article be $Q$, that of the second $R$, and the apparent and true weights of the third article $X$ and $W$ respectively. Let $a$ and $b$ be the lengths of the arms, $S$ and $S'$ the weights of the pans, $w$ the weight of the instrument, $x$ the distance of the centre of gravity from the fulcrum. Then we have the equations
$$(Q+S)a + w \cdot x = (Q_1+S')b,$$
$$(Q_2+S)a + w \cdot x = (Q+S')b,$$
$$(W+S)a + w \cdot x = (X+S')b,$$
$$(X+S)a + w \cdot x = (W+S')b,$$
whence $(W-Q_2) = (X-Q)b, \;\;\; (X-Q) = (W-Q_1)b,$
therefore $\frac{W-Q_2}{W-Q_1} = \frac{b^2}{a^2} = \frac{W-R_2}{W-R_1}$ similarly.
This gives the required value of W.
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Let $ABCD$ be the square, $A$ the corner of the square in contact with the wall, $B$ the corner to which the string is attached. Draw $BN$ perpendicular to the wall, and let the direction of the string meet the horizontal through $A$ in $O$. Then $O$ must be vertically below the centre of the square. Let the distances of $B, C$ and $D$ from the wall be $x, y, z$ respectively. Then $AO = 2x$.
$\;\;\;$ Also $\frac{y}{2} = \frac{x+z}{2} =$ distance of centre from wall $= 2x,$ whence $\frac{x}{1} = \frac{y}{4} = \frac{z}{3}$
and evidently $AN = z,$ so that the angle required is $\cot^{-1}\frac{z}{x} = \cot^{-1}3.$
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When the wedge is at rest, let $T$ be the tension of the string, $R$ the reaction between the plane and the mass $m'$. Then, if $f$ is the acceleration,
$$T-m'g\sin\alpha = m'f,\;\; mg-T = mf,$$
whence
$$T = \frac{mm'(1+\sin\alpha)}{m+m'} \cdot g.$$
Also $R = m'g \cos\alpha,$ and the horizontal force on the wedge is
$$R\sin\alpha - T\cos\alpha = \frac{m'cos\alpha (m'\sin\alpha - m)}{m+m'} \cdot g.$$
and to keep the wedge from moving, a force equal and opposite to this is required.
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If $\alpha$ is the angle of projection, and $\beta$ the angle of the plane, the vertical altitude above the plane after time $t$ is
$$ u\sin\alpha \cdot t - \frac12gt^2 - (u\cos\alpha \cdot t)\tan\beta = ut \cdot \frac{\sin(\alpha-\beta)}{\cos\beta} - \frac12gt^2.$$
Now, whatever $\lambda$ may be,
$$ ut \cdot \lambda - \frac12gt^2 = \frac12 \cdot \frac{u^2 \lambda^2}{g} - \frac12g(\frac{u\lambda}{g}-t)^2,$$
and therefore its greatest value is $\frac12 \cdot \frac{u^2 \lambda^2}{g}.$
$\;\;\;$ Hence the greatest value of the above vertical altitude is
$$ \frac12 \cdot \frac{u^2 \sin^2 (\alpha-\beta)}{g \cos^2 \beta} = \frac14 \cdot \frac{u^2}{g \cos^2 \beta} \left(1-\cos2(\alpha-\beta)\right).$$
But for the maximum range $2\alpha-\beta = \frac{\pi}{2}.$ Hence the above is
$$ \frac14 \cdot \frac{u^2}{g \cos^2 \beta} (1-\sin\beta),$$
i.e. $\frac14$ of the maximum range.
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