United Kingdom Mathematics Trust
英国数学信托(United Kingdom Mathematics Trust)
Senior Mathematical Challenge Tuesday 4 October 2022
高级数学挑战赛 2022年10月4日(星期二)
supported by \( \left\lbrack {\mathrm{{XTX}}\_ \mathrm{S}\text{ verleaf }}\right\rbrack \)
由\( \left\lbrack {\mathrm{{XTX}}\_ \mathrm{S}\text{ verleaf }}\right\rbrack \)支持
For reasons of space, these solutions are necessarily brief.
受篇幅限制，这些解答必然简洁。
There are more in-depth, extended solutions available on the UKMT website,
英国数学信托(UKMT)网站提供更深入、更详尽的解答，
which include some exercises for further investigation.
其中还包含一些供进一步探究的练习。
There is also a version of this document available on the UKMT website
英国数学信托(UKMT)网站还提供本文件的另一版本，
which includes each of the questions alongside its solution: www.ukmt.org.uk
其中每道题与其解答并列呈现:www.ukmt.org.uk
1.
D The expression simplifies to \( \frac{3 \times 8 \times {15} \times {24}}{\left( {2 \times 3}\right) \times \left( {3 \times 4}\right) \times \left( {4 \times 5}\right) \times \left( {5 \times 6}\right) } \) . Cancelling common factors gives \( \frac{1}{5} \) .
1.
D 该表达式化简为\( \frac{3 \times 8 \times {15} \times {24}}{\left( {2 \times 3}\right) \times \left( {3 \times 4}\right) \times \left( {4 \times 5}\right) \times \left( {5 \times 6}\right) } \)。约去公因数后得到\( \frac{1}{5} \)。
2.
C For the sum of five different primes to be prime, each of those five primes must be odd, Listing the primes starting with \( 3,5,7,{11},{13},{17},{19},\ldots \) and working systematically through possible sums gives a smallest sum of \( 3 + 5 + 7 + {11} + {13} = {39} \) which is not prime. However, the next smallest sum \( 3 + 5 + 7 + {11} + {17} = {43} \) which is prime as required.
2.
C 要使五个不同质数之和仍为质数，这五个质数必须全为奇数。从\( 3,5,7,{11},{13},{17},{19},\ldots \)开始依次列出质数并系统地尝试所有可能的和，得到的最小和为\( 3 + 5 + 7 + {11} + {13} = {39} \)，但此和并非质数。然而，下一个更小的和\( 3 + 5 + 7 + {11} + {17} = {43} \)恰为质数，满足题意。
3.
C Each of the possible parallelograms is formed from two adjacent equilateral triangles: \( P \) and \( Q, Q \) and \( R, R \) and \( S, S \) and \( T, T \) and \( U \) and finally \( U \) and \( P \) . Therefore there are six possible parallelograms.
3.
C 每一个可能的平行四边形均由两个相邻的等边三角形构成:\( P \)与\( Q, Q \)、\( R, R \)与\( S, S \)、\( T, T \)与\( U \)，以及最后\( U \)与\( P \)。因此共有六个可能的平行四边形。
4.
\( \mathbf{D} \) The area of the small square is \( 2 \times 2 = 4 \) . The area of the shaded region is then \( \frac{1}{2} \times 5 \times 5 - \frac{1}{2} \times 2 \times 2 \) \( = \frac{{25} - 4}{2} = \frac{21}{2} \) . Therefore the ratio of the area of the small square to the area of the shaded region is \( 4 : \frac{21}{2} = 8 : {21} \)
4.
\( \mathbf{D} \) 小正方形的面积为\( 2 \times 2 = 4 \)。于是阴影区域的面积为\( \frac{1}{2} \times 5 \times 5 - \frac{1}{2} \times 2 \times 2 \)\( = \frac{{25} - 4}{2} = \frac{21}{2} \)。因此小正方形面积与阴影区域面积之比为\( 4 : \frac{21}{2} = 8 : {21} \)。
5.
C Rewriting the calculation as \( \frac{100}{101} + \frac{10}{11} + \frac{1}{1} + \frac{1}{11} + \frac{1}{101} \) shows that we can reorder the sum to give \( \frac{101}{101} + \frac{11}{11} + \frac{1}{1} = 3 \)
5.
C 将计算改写为\( \frac{100}{101} + \frac{10}{11} + \frac{1}{1} + \frac{1}{11} + \frac{1}{101} \)，可见可将求和顺序重排为\( \frac{101}{101} + \frac{11}{11} + \frac{1}{1} = 3 \)。
6.
E Rewriting \( \frac{{4}^{800}}{{8}^{400}} \) using a base of 2 gives \( \frac{{\left( {2}^{2}\right) }^{800}}{{\left( {2}^{3}\right) }^{400}} = \frac{{2}^{1600}}{{2}^{1200}} = {2}^{400} \) using rules of indices.
6.
E 利用指数法则，将\( \frac{{4}^{800}}{{8}^{400}} \)改写为以2为底，可得\( \frac{{\left( {2}^{2}\right) }^{800}}{{\left( {2}^{3}\right) }^{400}} = \frac{{2}^{1600}}{{2}^{1200}} = {2}^{400} \)。
7.
C Consider first the units digits of 85 and 66. Multiples of 5 can only end in 5 or 0 . No multiples of 6 , an even number, can end in 5 . So in order that the units digit of our sum can be 0 , each of the multiples of 85 and 66 must individually have units digits of 0 . The smallest multiples of 85 and 66 with this property, \( 2 \times {85} = {170} \) and \( 5 \times {66} = {330} \) , have sum 500 . So 7 is the smallest number of stamps and they cost \( \pounds 5 \) .
7.
C 首先考虑85和66的个位数。5的倍数个位只能是5或0；6为偶数，其倍数个位不可能是5。因此要使和的个位为0，85和66各自的倍数个位必须为0。满足此条件的最小倍数分别为\( 2 \times {85} = {170} \)和\( 5 \times {66} = {330} \)，其和为500。故所需邮票的最小数量为7，总费用为\( \pounds 5 \)。
8.
E By drawing extra lines from the centre of the outer hexagon to each of its vertices and from the centre to the midpoint of each edge of the outer hexagon, 12 in total, the diagram can be shown to be made of 36 congruent triangles each with angles \( {30}^{ \circ },{60}^{ \circ } \) and \( {90}^{ \circ } \) . Twelve of these triangles are shaded giving a shaded area of \( \frac{1}{3} \times {216} = {72} \) .
8.
E 通过从外正六边形中心向每个顶点以及每条边的中点各作一条辅助线，共12条，可将图形划分为36个全等三角形，每个三角形的角分别为\( {30}^{ \circ },{60}^{ \circ } \)和\( {90}^{ \circ } \)。其中12个三角形被阴影覆盖，因此阴影部分面积为\( \frac{1}{3} \times {216} = {72} \)。
9.
B Using speed \( = \frac{\text{ distance }}{\text{ time }} \) gives \( 3 \times {10}^{8} = \frac{d}{{10}^{-9}} \) . Therefore \( d = 3 \times {10}^{-1}\mathrm{\;m} = {0.3}\mathrm{\;m} = {30}\mathrm{\;{cm}} \) .
9.
B 使用速度\( = \frac{\text{ distance }}{\text{ time }} \)得到\( 3 \times {10}^{8} = \frac{d}{{10}^{-9}} \)。因此\( d = 3 \times {10}^{-1}\mathrm{\;m} = {0.3}\mathrm{\;m} = {30}\mathrm{\;{cm}} \)。
10.
\( \mathbf{D} \) Rearranging the equation gives \( 1 + {2x} + 3{x}^{2} = 9 + {6x} + 3{x}^{2} \) so \( 1 + {2x} = 9 + {6x} \) and \( {4x} = - 8 \) . Therefore \( x = - 2 \) .
10.
\( \mathbf{D} \) 整理方程得\( 1 + {2x} + 3{x}^{2} = 9 + {6x} + 3{x}^{2} \)，所以\( 1 + {2x} = 9 + {6x} \)且\( {4x} = - 8 \)。因此\( x = - 2 \)。
11.
A When expressed as the product of its prime factors, \( {2022} = 2 \times 3 \times {337} \) . However, the integer \( n \) must be a factor of each integer in the middle row and so \( {n}^{2} \) must be a factor of their product 2022. Therefore \( n = 1 \) .
11.
A 当表示为质因数乘积时，\( {2022} = 2 \times 3 \times {337} \)。然而，整数\( n \)必须是中间行每个整数的因数，因此\( {n}^{2} \)必须是它们乘积2022的因数。因此\( n = 1 \)。
12.
E Using the difference of two squares, the calculation we are given can be written in the form \( {6666666}^{2} - {3333333}^{2} = \left( {{6666666} + {3333333}}\right) \left( {{6666666} - {3333333}}\right) = {9999999} \times {3333333} = \) \( {10000000} \times {3333333} - 1 \times {3333333} = {333333330000000} - {3333333} = {33333322666667} \) . The sum of the digits of this integer is 63 .
12.
E 利用平方差公式，给定的计算可写成形式\( {6666666}^{2} - {3333333}^{2} = \left( {{6666666} + {3333333}}\right) \left( {{6666666} - {3333333}}\right) = {9999999} \times {3333333} = \)\( {10000000} \times {3333333} - 1 \times {3333333} = {333333330000000} - {3333333} = {33333322666667} \)。该整数的各位数字之和为63。
13.
C Let the area of floor covered by exactly one rug be \( a \) , the area of floor covered by exactly two rugs be \( b \) and the area of floor covered by three rugs be \( c \) . Therefore, \( a + {2b} + {3c} = {90} \) and \( a + b + c = {60} \) . Subtracting the second equation from the first leaves \( b + {2c} = {30} \) and using \( b = {12} \) gives \( c = 9 \) .
13.
C 设恰好被一张地毯覆盖的地板面积为\( a \)，恰好被两张地毯覆盖的面积为\( b \)，被三张地毯覆盖的面积为\( c \)。因此\( a + {2b} + {3c} = {90} \)且\( a + b + c = {60} \)。将第二个方程从第一个方程中减去得\( b + {2c} = {30} \)，利用\( b = {12} \)得\( c = 9 \)。
14.
A Let \( {KL} \) be 3 units long. Then \( {KP} = 1,{PL} = 2 \) and area \( {KLMN} = \) \( 3 \times 3 = 9 \) . Removing four right-angled triangles congruent to \( {PLQ} \) from square \( {KLMN} \) gives area \( {PQRS} = 9 - 4 \times \frac{1}{2} \times 1 \times 2 = 5 \) . The area of \( {PQRS} \) is \( \frac{5}{9} \) of the area of \( {KLMN} \) . By the same reasoning the area of \( {TUVW} \) is \( \frac{5}{9} \) of the area of \( {PQRS} \) .
14.
A 设\( {KL} \)长度为3单位。则\( {KP} = 1,{PL} = 2 \)且面积\( {KLMN} = \)\( 3 \times 3 = 9 \)。从正方形\( {KLMN} \)中移除四个与\( {PLQ} \)全等的直角三角形，得到面积\( {PQRS} = 9 - 4 \times \frac{1}{2} \times 1 \times 2 = 5 \)。\( {PQRS} \)的面积是\( {KLMN} \)面积的\( \frac{5}{9} \)。同理，\( {TUVW} \)的面积是\( {PQRS} \)面积的\( \frac{5}{9} \)。
Combining these proportions gives the shaded area as \( \frac{5}{9} \times \frac{5}{9} = \frac{25}{81} \) of the area of \( {KLMN} \) .
将这些比例合并，阴影部分面积为\( {KLMN} \)面积的\( \frac{5}{9} \times \frac{5}{9} = \frac{25}{81} \)。
15.
B When the hare and tortoise are moving in the same direction, the hare completes \( {100}\mathrm{\;m} \) while the tortoise completes \( {25}\mathrm{\;m} \) . After the hare reverses direction and the hare and tortoise are moving towards one another, the hare is still moving four times as fast.
15.
B 当兔子和乌龟同向移动时，兔子完成\( {100}\mathrm{\;m} \)，乌龟完成\( {25}\mathrm{\;m} \)。兔子调头后，两者相向而行，兔子速度仍是乌龟的四倍。
Therefore the meeting point, \( M \) , is \( \frac{4}{5} \) of \( {75}\mathrm{\;m} = {60}\mathrm{\;m} \) away from the finish line.
因此相遇点\( M \)距离终点线\( {75}\mathrm{\;m} = {60}\mathrm{\;m} \)的\( \frac{4}{5} \)。
16.
B As \( x \) and \( y \) are interchangeable in the equation, the graph must be symmetric about the line \( y = x \) . This excludes options \( C \) and \( D \) . Substituting \( x = 0 \) and \( x = 1 \) into the equation shows that the graph crosses the axes at(0,1)and(1,0). Note that in option \( \mathrm{E} \) the line \( x + y = 1 \) meets \( y = x \) at \( \left( {\frac{1}{2},\frac{1}{2}}\right) \) whereas our curve meets \( y = x \) at \( \left( {\frac{1}{4},\frac{1}{4}}\right) \) and must therefore lie below the straight line shown in option E. The only possible option then is B.
16.
B 由于\( x \)与\( y \)在方程中可互换，图像必关于直线\( y = x \)对称，从而排除选项\( C \)和\( D \)。将\( x = 0 \)与\( x = 1 \)代入方程可知图像在(0,1)与(1,0)处穿过坐标轴。注意在选项\( \mathrm{E} \)中，直线\( x + y = 1 \)与\( y = x \)交于\( \left( {\frac{1}{2},\frac{1}{2}}\right) \)，而我们的曲线与\( y = x \)交于\( \left( {\frac{1}{4},\frac{1}{4}}\right) \)，因此必位于选项E所示直线下方，唯一可能的选项为B。
17.
B We enclose the regular octagon within a square as shown. Since the side-length of the octagon is 1 , the right-angled isosceles triangles in the corners have two short sides of length \( \frac{\sqrt{2}}{2} \) and so the square has side-length \( 1 + \sqrt{2} \) . Each of the right-angled triangles has area \( \frac{1}{2} \times \frac{\sqrt{2}}{2} \times \frac{\sqrt{2}}{2} = \frac{1}{4} \) . Each of the equilateral triangles which were removed has base 1 and so height \( \frac{\sqrt{3}}{2} \) . The shaded area can be obtained as the area of the square minus that of the four isosceles corners and the four equilateral triangles; that is \( {\left( 1 + \sqrt{2}\right) }^{2} - 4 \times \frac{1}{4} - 4 \times \frac{1}{2} \times \frac{\sqrt{3}}{2} = 3 + 2\sqrt{2} - 1 - \sqrt{3} = 2 + 2\sqrt{2} - \sqrt{3} \) .
17.
B 我们将正八边形按图示放入正方形内。因八边形边长为1，四个角上的等腰直角三角形的两直角边长均为\( \frac{\sqrt{2}}{2} \)，故正方形边长为\( 1 + \sqrt{2} \)。每个直角三角形面积为\( \frac{1}{2} \times \frac{\sqrt{2}}{2} \times \frac{\sqrt{2}}{2} = \frac{1}{4} \)。被移除的每个等边三角形底为1，故高为\( \frac{\sqrt{3}}{2} \)。阴影面积等于正方形面积减去四个等腰直角三角形与四个等边三角形的面积，即\( {\left( 1 + \sqrt{2}\right) }^{2} - 4 \times \frac{1}{4} - 4 \times \frac{1}{2} \times \frac{\sqrt{3}}{2} = 3 + 2\sqrt{2} - 1 - \sqrt{3} = 2 + 2\sqrt{2} - \sqrt{3} \)。
18.
B Let \( {3}^{x} = X \) and \( {3}^{y} = Y \) . The two equations can then be written as \( X + {3Y} = 5\sqrt{3} \) and \( {3X} + Y = 3\sqrt{3} \) . Subtracting three lots of the second equation from the first gives \( - {8X} = - 4\sqrt{3} \) so \( X = \frac{\sqrt{3}}{2} \) . Subtracting three lots of the first equation from the second gives \( - {8Y} = - {12}\sqrt{3} \) so \( Y = \frac{3\sqrt{3}}{2} \) . The value of \( {3}^{x} + {3}^{y} = X + Y = \frac{\sqrt{3}}{2} + \frac{3\sqrt{3}}{2} = 2\sqrt{3} \) . Alternatively, we could add the two equations giving \( {4X} + {4Y} = 8\sqrt{3} \) . Dividing by \( 4, X + Y = {3}^{x} + {3}^{y} = 2\sqrt{3} \) without knowing the value of either \( {3}^{x} \) or \( {3}^{y} \) individually.
18.
B 设\( {3}^{x} = X \)与\( {3}^{y} = Y \)，则两方程可写为\( X + {3Y} = 5\sqrt{3} \)与\( {3X} + Y = 3\sqrt{3} \)。将第二方程的三倍从第一方程中减去得\( - {8X} = - 4\sqrt{3} \)，故\( X = \frac{\sqrt{3}}{2} \)。将第一方程的三倍从第二方程中减去得\( - {8Y} = - {12}\sqrt{3} \)，故\( Y = \frac{3\sqrt{3}}{2} \)。\( {3}^{x} + {3}^{y} = X + Y = \frac{\sqrt{3}}{2} + \frac{3\sqrt{3}}{2} = 2\sqrt{3} \)的值。亦可两方程相加得\( {4X} + {4Y} = 8\sqrt{3} \)，再除以\( 4, X + Y = {3}^{x} + {3}^{y} = 2\sqrt{3} \)，而无需单独知道\( {3}^{x} \)或\( {3}^{y} \)的值。
19.
E The first equation can be rearranged to the form \( y = {x}^{2} - {2022} \) which is a translation of \( y = {x}^{2} \) down 2022 units. The second equation is a reflection of the first, in the line \( y = x \) . There are four points of intersection of these two parabolas.
19.
E 第一方程可整理为\( y = {x}^{2} - {2022} \)，即把\( y = {x}^{2} \)向下平移2022个单位。第二方程是第一方程关于直线\( y = x \)的反射。这两条抛物线共有四个交点。
20.
D Let \( O \) be the centre of the circle, \( M \) and \( N \) be the midpoints of two sides of the square and \( V \) and \( P \) be vertices of two sides of the square as shown. Line \( {ONM} \) is a line of symmetry. Let \( {ON} = x \) . Therefore \( {NP} = x \) as \( {ON} \) and \( {NP} \) are sides of the right-angled isosceles triangle \( {ONP} \) . Also, \( {PV} = {MN} = {2x} \) . Consider right-angled triangle \( {OVM} \) . The radius of the circle is given as 10, therefore \( {\left( 3x\right) }^{2} + {x}^{2} = {10}^{2} \) so \( {10}{x}^{2} = {100} \) and \( {x}^{2} = {10} \) . Hence the area of the square is \( {\left( 2x\right) }^{2} = 4{x}^{2} = 4 \times {10} = {40} \) .
20.
D 设\( O \)为圆心，\( M \)和\( N \)为正方形两条边的中点，\( V \)和\( P \)为正方形两条边的顶点，如图所示。直线\( {ONM} \)为对称轴。设\( {ON} = x \)。因此\( {NP} = x \)，因为\( {ON} \)和\( {NP} \)是直角等腰三角形\( {ONP} \)的两条边。同时，\( {PV} = {MN} = {2x} \)。考虑直角三角形\( {OVM} \)。已知圆的半径为10，因此\( {\left( 3x\right) }^{2} + {x}^{2} = {10}^{2} \)，于是\( {10}{x}^{2} = {100} \)且\( {x}^{2} = {10} \)。故正方形的面积为\( {\left( 2x\right) }^{2} = 4{x}^{2} = 4 \times {10} = {40} \)。
21.
D Half the diagram is shown here. In it, the shaded area equals the area of a right-angled isosceles triangle of side-length 2 plus the area of a large semicircle minus the area of a small semicircle of radius 1. Using Pythagoras’ Theorem, the diameter of the large semicircle has length \( 2\sqrt{2} \) and so the radius is \( \sqrt{2} \) . Therefore the shaded area of the full diagram is \( 2\left\lbrack {\frac{1}{2} \times 2 \times 2 + \frac{1}{2}\pi \times {\left( \sqrt{2}\right) }^{2} - \frac{1}{2}\pi \times {1}^{2}}\right\rbrack = 2\left( {2 + \pi - \frac{1}{2}\pi }\right) = 4 + \pi . \)
21.
D 这里展示了图形的一半。其中，阴影部分的面积等于边长为2的直角等腰三角形的面积，加上大半圆的面积，再减去半径为1的小半圆的面积。根据勾股定理(Pythagoras’ Theorem)，大半圆的直径长度为\( 2\sqrt{2} \)，因此半径为\( \sqrt{2} \)。于是整个图形的阴影面积为\( 2\left\lbrack {\frac{1}{2} \times 2 \times 2 + \frac{1}{2}\pi \times {\left( \sqrt{2}\right) }^{2} - \frac{1}{2}\pi \times {1}^{2}}\right\rbrack = 2\left( {2 + \pi - \frac{1}{2}\pi }\right) = 4 + \pi . \)。
22.
B Squaring both sides of the equation gives \( x - \sqrt{x + {23}} = 8 - 4\sqrt{2}y + {y}^{2} \) which can be rearranged to \( \sqrt{x + {23}} = \left( {x - 8 - {y}^{2}}\right) + 4\sqrt{2}y\left\lbrack 1\right\rbrack \) . Squaring equation [1] gives \( x + {23} = {\left( x - 8 - {y}^{2}\right) }^{2} + \) \( 2\left( {4\sqrt{2}y}\right) \left( {x - 8 - {y}^{2}}\right) + {32}{y}^{2}\left\lbrack 2\right\rbrack \) . We are given that both \( x \) and \( y \) are integers and so the surd component, \( 2\left( {4\sqrt{2}y}\right) \left( {x - 8 - {y}^{2}}\right) \) , must equal 0 . Therefore either \( y = 0 \) or \( \left( {x - 8 - {y}^{2}}\right) = 0 \) [3]. Consider first the case \( y = 0 \) . Here, equation [2] reduces to \( x + {23} = {\left( x - 8\right) }^{2} \) . This expands to \( {x}^{2} - {17x} + {41} = 0 \) which has no integer solutions as its discriminant is \( {\left( -{17}\right) }^{2} - 4 \times 1 \times {41} = {125} \) , which is not square. Secondly considering \( \left( {x - 8 - {y}^{2}}\right) = 0\left\lbrack 3\right\rbrack \) reduces [1] to \( \sqrt{x + {23}} = 4\sqrt{2}y \) and therefore \( x + {23} = {32}{y}^{2} \) . Using [3] again gives \( x = 8 + {y}^{2} \) and so \( {31} + {y}^{2} = {32}{y}^{2} \) . Therefore \( {y}^{2} = 1 \) . Hence \( y = \pm 1 \) and in either case, \( x = 8 + 1 = 9 \) . Because equations have been squared, some solutions could be spurious. Substituting in the original equation, we see that(9,1)is a solution but(9, - 1)is not. Hence there is just one solution.
22.
B 将方程两边平方得到\( x - \sqrt{x + {23}} = 8 - 4\sqrt{2}y + {y}^{2} \)，可整理为\( \sqrt{x + {23}} = \left( {x - 8 - {y}^{2}}\right) + 4\sqrt{2}y\left\lbrack 1\right\rbrack \)。将方程[1]平方得\( x + {23} = {\left( x - 8 - {y}^{2}\right) }^{2} + \)\( 2\left( {4\sqrt{2}y}\right) \left( {x - 8 - {y}^{2}}\right) + {32}{y}^{2}\left\lbrack 2\right\rbrack \)。已知\( x \)和\( y \)均为整数，因此根式部分\( 2\left( {4\sqrt{2}y}\right) \left( {x - 8 - {y}^{2}}\right) \)必须为零。于是要么\( y = 0 \)，要么\( \left( {x - 8 - {y}^{2}}\right) = 0 \)[3]。先考虑\( y = 0 \)的情况。此时方程[2]简化为\( x + {23} = {\left( x - 8\right) }^{2} \)，展开得\( {x}^{2} - {17x} + {41} = 0 \)，其判别式为\( {\left( -{17}\right) }^{2} - 4 \times 1 \times {41} = {125} \)，不是完全平方，故无整数解。其次考虑\( \left( {x - 8 - {y}^{2}}\right) = 0\left\lbrack 3\right\rbrack \)，则[1]简化为\( \sqrt{x + {23}} = 4\sqrt{2}y \)，从而\( x + {23} = {32}{y}^{2} \)。再次利用[3]得\( x = 8 + {y}^{2} \)，于是\( {31} + {y}^{2} = {32}{y}^{2} \)。因此\( {y}^{2} = 1 \)。故\( y = \pm 1 \)，且无论哪种情况，\( x = 8 + 1 = 9 \)。由于方程曾被平方，部分解可能为增根。代回原方程验证，(9,1)是解，而(9,-1)不是。故仅有一组解。
23.
A The lengths of the sides of the three squares are \( \sqrt{10} \) , \( 3\sqrt{10} \) and \( 2\sqrt{10} \) respectively. Therefore \( {HQ} = 2\sqrt{10} \) and \( {RJ} = \sqrt{10} \) . In triangle \( {GQH} \) , the gradient of \( {GH} \) is \( \frac{2\sqrt{10}}{\sqrt{10}} = 2 \) . In triangle \( {JRK} \) , the gradient of \( {JK} \) is \( \frac{-\sqrt{1}0}{2\sqrt{10}} = \frac{-1}{2} \) . Therefore lines \( {FI} \) (on which \( {GH} \) lies) and \( {IL} \) (on which \( {JK} \) lies) are perpendicular.
23.
A 三个正方形的边长分别为\( \sqrt{10} \)、\( 3\sqrt{10} \)和\( 2\sqrt{10} \)，因此\( {HQ} = 2\sqrt{10} \)且\( {RJ} = \sqrt{10} \)。在三角形\( {GQH} \)中，\( {GH} \)的斜率为\( \frac{2\sqrt{10}}{\sqrt{10}} = 2 \)；在三角形\( {JRK} \)中，\( {JK} \)的斜率为\( \frac{-\sqrt{1}0}{2\sqrt{10}} = \frac{-1}{2} \)。因此直线\( {FI} \)(\( {GH} \)所在直线)与\( {IL} \)(\( {JK} \)所在直线)互相垂直。
All five right-angled triangles around the edge of the figure and triangle \( {FIL} \) itself are similar as they contain the same angles. They all have sides in the ratio \( 1 : 2 : \sqrt{5} \) . To calculate the area of triangle \( {FIL} \) we need the length \( {IL} \) , as the area of \( {FIL} = \frac{1}{2} \times {IL} \times \frac{1}{2}{IL} \) . The length \( {IL} \) is made of three sections: \( {JK} = \sqrt{10} \times \sqrt{5},{KL} = {2JK} = 2 \times \sqrt{10} \times \sqrt{5} \) and \( {IJ} = \frac{2}{\sqrt{5}} \times {HJ} = \frac{2}{\sqrt{5}} \times 3\sqrt{10} \) . Therefore \( {IL} = {IJ} + {JK} + {KL} = 6\sqrt{2} + \sqrt{50} + 2\sqrt{50} = {21}\sqrt{2} \) . Hence the area of triangle \( {FIL} = \frac{1}{2} \times {21}\sqrt{2} \times \frac{21}{2}\sqrt{2} = {220.5}. \)
图形边缘的所有五个直角三角形以及三角形\( {FIL} \)本身都是相似的，因为它们具有相同的角度。它们的边长比例均为\( 1 : 2 : \sqrt{5} \)。要计算三角形\( {FIL} \)的面积，我们需要长度\( {IL} \)，因为\( {FIL} = \frac{1}{2} \times {IL} \times \frac{1}{2}{IL} \)的面积如此。长度\( {IL} \)由三部分组成:\( {JK} = \sqrt{10} \times \sqrt{5},{KL} = {2JK} = 2 \times \sqrt{10} \times \sqrt{5} \)和\( {IJ} = \frac{2}{\sqrt{5}} \times {HJ} = \frac{2}{\sqrt{5}} \times 3\sqrt{10} \)。因此\( {IL} = {IJ} + {JK} + {KL} = 6\sqrt{2} + \sqrt{50} + 2\sqrt{50} = {21}\sqrt{2} \)。故三角形\( {FIL} = \frac{1}{2} \times {21}\sqrt{2} \times \frac{21}{2}\sqrt{2} = {220.5}. \)的面积为
24.
D Rearranging \( {xy} = {px} + {qy} \) to make \( y \) the subject, gives \( {xy} - {qy} = {px} \) so \( y\left( {x - q}\right) = {px} \) and therefore \( y = \frac{px}{x - q} \) which rearranges to \( y = p + \frac{pq}{x - q} \) . A sketch of the graph of this function for real values of \( x \) and \( y \) is shown. As \( x \) and \( y \) are both integers in this question, \( y \) takes its maximum value when \( x - q \) is as small as possible therefore \( x - q = 1 \) so \( x = q + 1 \) . The expression \( y - x \) then becomes \( \frac{px}{1} - x = \left( {p - 1}\right) x = \left( {p - 1}\right) \left( {q + 1}\right) \) . 25. A
24.
D 将\( {xy} = {px} + {qy} \)整理，使\( y \)成为主项，得到\( {xy} - {qy} = {px} \)，于是\( y\left( {x - q}\right) = {px} \)，因此\( y = \frac{px}{x - q} \)，整理得\( y = p + \frac{pq}{x - q} \)。图中给出了该函数在实数\( x \)和\( y \)下的图像草图。由于本题中\( x \)和\( y \)均为整数，当\( x - q \)尽可能小时，\( y \)取最大值，因此\( x - q = 1 \)，于是\( x = q + 1 \)。表达式\( y - x \)则变为\( \frac{px}{1} - x = \left( {p - 1}\right) x = \left( {p - 1}\right) \left( {q + 1}\right) \)。25. A
Let \( M \) be the midpoint of \( {QP} \) . The volume of the carton is \( \frac{1}{3} \times \) base area of triangle \( {PQS} \times \) the perpendicular height from \( \mathrm{R} \) to the plane containing PQS. Triangle PQS is isosceles and \( {MS} = \sqrt{{10}^{2} - {2}^{2}} = \sqrt{96} \) . So area of \( {PQS} = \frac{1}{2} \times 4 \times \sqrt{96} = 8\sqrt{6} \) . Consider isosceles triangle
设\( M \)为\( {QP} \)的中点。纸盒的体积为\( \frac{1}{3} \times \)三角形\( {PQS} \times \)的底面积乘以从\( \mathrm{R} \)到包含PQS的平面的垂直高。三角形PQS为等腰三角形，且\( {MS} = \sqrt{{10}^{2} - {2}^{2}} = \sqrt{96} \)。因此\( {PQS} = \frac{1}{2} \times 4 \times \sqrt{96} = 8\sqrt{6} \)的面积为。考虑等腰三角形
\( {MRS} \) and let \( N \) be the midpoint of \( {RS}.{MN} = \sqrt{{\left( \sqrt{96}\right) }^{2} - {2}^{2}} = \sqrt{92} \) , so with \( {RS} \) as the base, area of \( {MRS} = \frac{1}{2} \times 4 \times \sqrt{92} = 4\sqrt{23} \) . Now with \( {MS} \) as the base, area of \( {MRS} = \frac{1}{2} \times {MS} \times h \) . Therefore \( 4\sqrt{23} = \frac{1}{2} \times \sqrt{96} \times h \) and \( h = \frac{2\sqrt{23}}{\sqrt{6}} \) . Finally, the volume \( = \frac{1}{3} \times 8\sqrt{6} \times \frac{2\sqrt{23}}{\sqrt{6}} = \frac{{16}\sqrt{23}}{3} \) .
\( {MRS} \)，设\( N \)为\( {RS}.{MN} = \sqrt{{\left( \sqrt{96}\right) }^{2} - {2}^{2}} = \sqrt{92} \)的中点，于是以\( {RS} \)为底，\( {MRS} = \frac{1}{2} \times 4 \times \sqrt{92} = 4\sqrt{23} \)的面积为。再以\( {MS} \)为底，\( {MRS} = \frac{1}{2} \times {MS} \times h \)的面积为。因此\( 4\sqrt{23} = \frac{1}{2} \times \sqrt{96} \times h \)且\( h = \frac{2\sqrt{23}}{\sqrt{6}} \)。最终，体积\( = \frac{1}{3} \times 8\sqrt{6} \times \frac{2\sqrt{23}}{\sqrt{6}} = \frac{{16}\sqrt{23}}{3} \)。