เฉลยข้อสอบชุดที่ 1: MONDAY, December 31, 1877. 9:00 to 12:00
3. The opposite angles of any quadrilateral figure inscribed in a circle are together equal to two right angles.
$\;\;\;$ If a quadrilateral be inscribed in a circle, and the middle points of the arcs subtended by its sides be joined to make another quadrilateral, and so on : shew that these quadrilaterals tend to become squares.
Solution:
$\;\;\;$ Let $A_1B_1C_1D_1, \;A_2B_2C_2D_2, \;A_3B_3C_3D_3$ be three successive quadrilaterals, made as the question directs, the order of points being given by
$A_1A_2B_1B_2......\;$ and $\;A_2A_3B_2B_3......$
Then
$2$ arc $A_2B_2 =$ arc $A_1B_1 +$ arc $B_1C_1,$
$2$ arc $B_2C_2 =$ arc $B_1C_1 +$ arc $C_1D_1,$
$............ = ............ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$
$\;\;\;$ Let $A_1B_1$ be the greatest; $B_1C_1$ or $C_1D_1$ the least of the arcs $A_1B_1,\; B_1C_1,\; C_1D_1,\; D_1A_1$. Then the above equalities shew that the arc $A_2B_2$ is less than the greatest and greater than the least of these arcs, and that the same is true for the arcs $B_2C_2,\; C_2D_2,\; D_2A_2$. Hence $A_3B_3C_3D_3$ has no side so great and no side so small as the greatest and least sides of $A_1B_1C_1D_1$ respectively: i.e. $A_2B_2C_2D_2$ is more nearly equilateral than $A_1B_1C_1D_1$: and so on. Hence the quadrilaterals tend to become rhombi, but the only rhombus that can be inscribed in a circle is a square, therefore the quadrilaterals tend to become squares.
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