106上學期招生考試(試題+解答)r1

北區高中學生數學與科學跨領域研究人才培育計畫 考試時間: 106年9月24日, 13:30 – 16:30,計三小時 本試題共五題_,兩頁 試題若有疑問, 請於考試開始後的三十分鐘內, 舉手提交 「提問單」 詢問;之後不再接受詢 問。 A4 白紙為答案紙與計算紙, 考試結束請將答案排序, 然後提問單與計算紙排在最後 面, 再由監考人員裝訂。 答案限用黑色或藍色筆書寫, 僅作圖可使用鉛筆, 不得使用修正液 (帶), 不得使用電子計算器。 每題七分,答題的“推演過程”為評分的依據。 1. 定義函數m如下: m(x, y, z) = max{x2 , y2 , z2 }, x, y, z為實數, 此處_max{x2 , y2 , z2 }表示三數x2 , y2 , z2_{之最大值。 若實數} x, y, z滿足下列條件: x + y + z = 0, x2 + y2 + z2 = 1. 試求m的最小值。

Define the functionm of the three real variables x, y, z by

m(x, y, z) = max{x2

, y2

, z2

}, x, y, z ∈ ℜ, where ℜ is the set of real numbers. Determine, with proof, the minimum value of_{m if x, y, z vary in ℜ subject to the} following conditions: x + y + z = 0, x2 + y2 + z2 = 1. 2. 是否存在整數a與b,使得1062 _可以整除 a106 + b106 + 1?

Do integersa and b exist, such that a106

+ b106

+ 1 is divisible by 1062

?

3. 在三角形 ABC 中, 其內切圓分別與邊 BC, AC 相切於點D, E. 若邊 AD 與邊 BE 相等,試證:三角形ABC為等腰三角形。

In a triangleABC, the incircle touches the sides BC and AC at D and E respec-tively. Show that the triangle has to be isosceles if the lengths ofAD and BE are equal.

4. 伊薩克正在規劃他為期九天的假期。 假期中的每一天, 他都會去衝浪、 滑水或純休

息。 他每天只會做一件事, 且他不會有連續兩天進行 “不同”的水上活動。 伊薩克的

假期總共有幾種可能的安排?

Isaac is planning a nine-day holiday. Every day he will go surfing, or water skiing, or he will rest. On any given day he does just one of these three things. He never does different water-sports on consecutive days. How many schedules are possible for this holoday?

4.1 伊薩克正在規劃他為期九天的假期。 假期中的每一天, 他都會去衝浪、 滑水或純休 息。 他每天只會做一件事, 且他不會有連續兩天進行 “相同”的水上活動。 伊薩克的 假期總共有幾種可能的安排? 5. 令A為不是0之整數。試求下列聯立方程式之整數解: x + y2 + z3 = A, 1 x + 1 y2 + 1 z3 = 1 A, xy2 z3 = A2 . Let_{A 6= 0 be an integer.}

Find all integer solutions of the following system of equations:

x + y2 + z3 = A, 1 x + 1 y2 + 1 z3 = 1 A, xy2 z3 = A2 . 2

北區高中學生數學與科學跨領域研究人才培育計畫 參考解答 106年9月24日 1. 定義函數m如下: m(x, y, z) = max{x2 , y2 , z2 }, x, y, z為實數, 此處_max{x2 , y2 , z2 }表示三數x2 , y2 , z2_{之最大值。 若實數} x, y, z滿足下列條件: x + y + z = 0, x2 + y2 + z2 = 1. 試求m的最小值。

Define the functionm of the three real variables x, y, z by

m(x, y, z) = max{x2

, y2

, z2

}, x, y, z ∈ ℜ, where ℜ is the set of real numbers. Determine, with proof, the minimum value of_{m if x, y, z vary in ℜ subject to the} following conditions: x + y + z = 0, x2 + y2 + z2 = 1. 解: 不妨假設_{x, y ≥ 0 ≥ z.}由已知條件知:_{z = −(x + y)}且 2z2 = z2 + (x2 + 2xy + y2 ) = x2 + y2 + +z2 + 2xy = 1 + 2xy ≥ 1, 上式等號成立之充要條件為:x, y 其中一個為0. 換言之, m ≥ z2 ≥ 1 2, 上式等號成立之充要條件為:x, y 其中一個為0. 故m之最小值為1/2.

2. 是否存在整數a與b,使得1062 _可以整除

a106

+ b106

+ 1?

Do integersa and b exist, such that a106

+ b106

+ 1 is divisible by 1062

? 解: Since 106 is even, we know that

4|1062

.

Ifa and b with the given property exist, then

4|a106

+ b106

+ 1 and a106

+ b106

+ 1 is even.

This is only possible if one ofa and b is even and the other odd. Without loss of generality assumea odd and b even.

If _{b even and therefore 4|b}106

, and we have b106 ≡ 0 (mod 4). if a is odd, we either have a106 ≡ 1 (mod 4) or a106 ≡ 3 (mod 4). In either case, we have

a2

≡ 1 (mod 4), and therefore a106

≡ 1 (mod 4). It follows that

a106

+ b106

+ 1 ≡ 1 + 0 + 1 ≡ 2 (mod 4),

and this number can certainly not be divisible by 4, and therefore not by1062

3. 在三角形 ABC 中, 其內切圓分別與邊 BC, AC 相切於點D, E. 若邊 AD 與邊 BE 相等,試證:三角形ABC為等腰三角形。

In a triangleABC, the incircle touches the sides BC and AC at D and E respec-tively. Show that the triangle has to be isosceles if the lengths ofAD and BE are equal.

解: Consider the trianglesADC and BEC. Since CD and CE are tangents from C to the incircle, their lengths have to be the same. By the assumption, the lengths of AD and BE are equal as well. Furthermore, ∠ACD = ∠BCE, so the triangles ADC and BEC have two side lengths and one angle in common. Hence, in order to show that they are congruent, it is suficient to prove that the angles_{∠CAD and} ∠CBE are acute.

A B

C

D E

Since the length ofCD equals the length of CE and E lies on the segment AC, CD is shorter than AC, which means that ∠CAD < ∠CDA. Thus ∠CAD is acute, and by similar arguments,∠CBE is acute too.

Therefore, ∆ADC ≅ ∆BEC, and so the side lengths AC and BC are equal, which finishes the proof.

4. 伊薩克正在規劃他為期九天的假期。 假期中的每一天, 他都會去衝浪、 滑水或純休

息。 他每天只會做一件事, 且他不會有連續兩天進行 “不同”的水上活動。 伊薩克的

假期總共有幾種可能的安排?

Isaac is planning a nine-day holiday. Every day he will go surfing, or water skiing, or he will rest. On any given day he does just one of these three things. He never does different water-sports on consecutive days. How many schedules are possible for this holoday?

解: Split the 9 days into three sets of 3. Writing “R” for resting, “S” for surfing and “W ” for waterskiing, there are 17 possible schedules for a three day stretch, namely

SSS, SSR, SRR, SRS, SRW,

W W W, W W R, W RR, W RW, W RS,

RRR, RRS, RSR, RSS, RRW, RW R, RW W.

How many ways can we piece three of these together? If we specify the last day of each of the first two sets of three, then we can calculate the number of possible ways - the restrictions are that if the first set ends in anS, then the second set must beginS or R, and if it ends in W , then the second set must begin W or R.

First set ends in_{S, second in S: 5 × 4 × 12 = 240.} First set ends in_{S, second in R: 5 × 5 × 17 = 425.} First set ends in_{S, second in W : 5 × 3 × 12 = 180.} First set ends in_{R, second in S: 7 × 5 × 12 = 420.} First set ends in_{R, second in R: 7 × 7 × 17 = 833.} First set ends in_{R, second in W : 7 × 5 × 12 = 420.} First set ends in_{W , second in S: 5 × 3 × 12 = 180.} First set ends in_{W , second in R: 5 × 5 × 17 = 425.} First set ends in_{W , second in W : 5 × 4 × 12 = 240.}

The total number of possible schedules is the sum of the nine numbers,

4.1 伊薩克正在規劃他為期九天的假期。 假期中的每一天, 他都會去衝浪、 滑水或純休 息。 他每天只會做一件事, 且他不會有連續兩天進行 “相同”的水上活動。 伊薩克的 假期總共有幾種可能的安排? 解: 令S(n), W (n), R(n)分別為在給定第n 天為 衝浪(S),滑水(W), 休息(R)的前提 下,前n天的可能排列數,則我們有: 1. S(1) = W (1) = R(1) = 1; 2. S(n + 1) = R(n) + W (n) (因為如果第n + 1天是S,則第n天不可能是S); 3. W (n + 1) = R(n) + S(n) (因為如果第n + 1天是 W ,則第n 天不可能是 W ); 4. R(n + 1) = S(n) + R(n) + W (n) (因為如果第n + 1天休息,第n天可以是 任何活動)。 由此我們可以如下計算出所有的值,最終得到第9天的所有可能為 985 + 985 + 1393 = 3363. n 1 2 3 4 5 6 7 8 9 S(n) 1 2 5 12 29 70 169 408 985 W (n) 1 2 5 12 29 70 169 408 985 R(n) 1 3 7 17 41 99 239 577 1393

5. 令A為不是0之整數。試求下列聯立方程式之整數解: x + y2 + z3 = A, 1 x + 1 y2 + 1 z3 = 1 A, xy2 z3 = A2 . Let_{A 6= 0 be an integer.}

Find all integer solutions of the following system of equations:

x + y2 + z3 = A, 1 x + 1 y2 + 1 z3 = 1 A, xy2 z3 = A2 .

解: We multiply the lase two equations and obtain y2

z3

+ xz3

+ xy2

= A. We now have the values of all elementary sysmetric functions of x, y2

, z3

and obtain the polynomial (t − x)(t − y2 )(t − z3 ) = t3 − At2 + At − A2 = (t − A)(t2 + A)

= (t − A)(t −√−A)(t +√−A). If_{A = −B}2for some positive integerB. We have to match x, y2

, z3

with_−B2_{, B, −B.} We must have y2

= B since this is the only nonnegative choice. Both the remaining choices demand thatB is a perfect cube in addition to being a perfect square, soB has to be of the form C6

.

In summary, there is no solution if _{A is not the form −C}12

for some positive integerC.

If_{A = −C}12

for some positive integerC, we have two solutions: (i) The case(x, y2

, z3 ) = (−B2 , B, −B) gives (x, y, z) = (−C12 , ±C3 , −C2 ).

(ii) The case(x, y2

, z3 ) = (−B, B, −B2 ) gives (x, y, z) = (−C6 , ±C3 , −C4 ).