北區高中學生數學與科學跨領域研究人才培育計畫(MiGhty) 105學年 上學期 期中評量 考題 考試時間: 105年11月19日,計四小時 本試題共六題。 個人總分依據得分最高的三題。 題目若細分子題則加總子題分數。 比序若 遇到同分時,則參考第四高分題。 試題若有疑問, 請於考試開始後的六十分鐘內, 舉手並提交 「提問單」詢問;之後不再接受 詢問。 請用A4白紙為答案紙與計算紙,考試結束請將答案排序,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。 答題的“推演過程”為評分依據,簡答沒有分數。 每題分數標示在題號之後。 於13:00之前 與最後5分鐘,不得提前交卷。 1. [5 pts]令n > 1為整數, 滿足於: 若d是n的因數,則d + 1也會是n + 1的因數。 證明: n是質數。
An integer n > 1 has the following property: for every positive divisor d of n, d + 1 is a divisor of n + 1. Prove that n is prime.
註:中英文題目是等價的,但是英文題目的“every”卻是中文題目沒有明明白白說出
來的重點。
2. [7 pts]令U V W Z為矩形,而A和B分別是邊ZW 和ZU 上的點。 從A對於BV 的垂線和直線V W 交於X, 從B 對於AV 的垂線和直線V U 交於Y . 證明:X, Y
和Z 三點共線。
Let U V W Z be a rectangle, and let A and B be points on sides ZW and ZU , respectively. The perpendicular from A to BV intersects line V W in X, and the perpendicular from B to AV intersects line V U in Y . Prove that X, Y and Z are collinear (lie on the same line).
3. [7 pts]求所有的實係數多項式P 滿足P (x2) = P (x)P (x + 2). (x也是實數。) Find all polynomials P with real coefficients satisfying P (x2) = P (x)P (x + 2). (x
is a real variable.)
4. [9 pts] 一場考試出了m 題四選一的選擇題, 有16個學生參加這場考試, 他們每一
題都有作答,且任兩人只有至多一題答案相同,請問m至多為多少?
A test consists of m multiple choice problems each of which offers four options to select the right one. There are 16 contestants. It is known that they answer every problem. And for any two contestants, their answer sheets have at most one problem chosen the same answer. What is the largest m?
5. [9 pts] 給定一個圓內接四邊形 ABCD, 令對角線 AC 和 BD 交在 P . 由 P 到 AB 和 CD 的垂足分別為 E 和 F , 而AD 和BC 的中點分別為 M 和N , 證明: EM F N 是一個箏形。
Given a cyclic quadrilateral ABCD. Let P be the intersection of the two diagonals AC and BD. The pedals of the perpendicular lines from P to AB and CD are respectively E and F , and the midpoints of AD and BC are respectively M and N . Prove that EM F N is a deltoid (kite shape).
6. [10 pts]令a1, a2, . . .為正實數數列,而且對任意n ≥ 1有下列不等式 a2n+1 ≥ a 2 1 13 + a22 23 + · · · + a2n n3. 試證:存在一個正整數M 使得 M X n=1 an+1 a1+ a2+ · · · + an > 2016 1016.
Let a1, a2, . . . be a sequence of positive real numbers such that for each n ≥ 1,
a2n+1 ≥ a 2 1 13 + a2 2 23 + · · · + a2 n n3.
Show that there is a positive integer M such that
M X n=1 an+1 a1+ a2+ · · · + an > 2016 1016. 2
105學年 北區高中學生數學與科學跨領域研究人才培育計畫(MiGhty) 上學期 期中評量 考題 參考解答
105年11月19日
1. [5 pts]令n > 1為整數, 滿足於: 若d是n的因數,則d + 1也會是n + 1的因數。 證明: n是質數。
An integer n > 1 has the following property: for every positive divisor d of n, d + 1 is a divisor of n + 1. Prove that n is prime.
註:中英文題目是等價的,但是英文題目的“every”卻是中文題目沒有明明白白說出
來的重點。
解: Assume that n is not a prime number. Let p be the smallest prime factor of n, and let d = n/p 6= 1. Then np + p n + p = p(n + 1) p(d + 1) = n + 1 d + 1
is an integer, by assumption. But n + p also divides np + p2, so it must divide the difference (np + p2) − (np + p) = p2− p, i.e., n + p | p2− p. This certainly gives
n + p ≤ p2− p, and then n = dp < p2. Dividing by p, we have d < p. Suppose that
d has some prime factor q; then q ≤ d < p. On the other hand, q also divides n, and then the minimality of p gives q ≥ p. This is a contradiction, so q cannot exist, and we conclude that d = 1. Then, n = p, as needed.
另一種解:假設n不是質數,可令n = a × b,其中a ≥ b > 1. 則 n + 1 a + 1 = ab + 1 a + 1 = b − b − 1 a + 1 然而a + 1 > b − 1 > 0,所以 b−1 a+1 非整數,也就是a + 1無法整除n + 1,與題意矛 盾。
2. [7 pts]令U V W Z為矩形,而A和B分別是邊ZW 和ZU 上的點。 從A對於BV 的垂線和直線V W 交於X, 從B 對於AV 的垂線和直線V U 交於Y . 證明:X, Y
和Z 三點共線。
Let U V W Z be a rectangle, and let A and B be points on sides ZW and ZU , respectively. The perpendicular from A to BV intersects line V W in X, and the perpendicular from B to AV intersects line V U in Y . Prove that X, Y and Z are collinear (lie on the same line).
解: Observe that∠XAW = ∠XV B = ∠V BU . Hence 4XAW ∼ 4V BU , and thus XW/V U = AW/BU . Likewise, Y U/V W = BU/AW . Putting these together yields XW/V U = V W/Y U , and hence
XW ZW =
ZU
Y U ⇒ 4XZW ∼ 4ZY U.
Since ∠UZW = 90◦, this immediately implies ∠Y ZX = 180◦, and then X, Y and Z are collinear.
3. [7 pts]求所有的實係數多項式P 滿足P (x2) = P (x)P (x + 2). (x也是實數。) Find all polynomials P with real coefficients satisfying P (x2) = P (x)P (x + 2). (x
is a real variable.)
解: 1. If P is a constant function:
Assume P ≡ c. Then the condition becomes c = c2. So c = 0 or 1.
2. If P is non-constant:
Suppose the degree of P is n. Let Q(x) = P (x) − (x − 1)n. Then
Q(x2) + (x2− 1)n = P (x2) = P (x)P (x + 2) = (Q(x) + (x − 1)n) (Q(x + 2) + (x + 1)n) = Q(x)Q(x + 2) + (x + 1)nQ(x) + (x − 1)nQ(x + 2) + (x2− 1)n, which is equivalent to Q(x2) = Q(x)Q(x + 2) + (x + 1)nQ(x) + (x − 1)nQ(x + 2).
If Q is non-zero, suppose the degree of Q is k. It’s easy to see that the degree of LHS is 2k and the degree of RHS is n + k. Notice that n must bigger than k. (Why?) So this situation can’t happen. In other words, Q is zero. That is P (x) = (x − 1)n
We conclude that the possibilities of P is 0, 1 and (x − 1)n, which are indeed the solutions.
4. [9 pts] 一場考試出了m 題四選一的選擇題, 有16個學生參加這場考試, 他們每一
題都有作答,且任兩人只有至多一題答案相同,請問m至多為多少?
A test consists of m multiple choice problems each of which offers four options to select the right one. There are 16 contestants. It is known that they answer every problem. And for any two contestants, their answer sheets have at most one problem chosen the same answer. What is the largest m?
解: 這個數m至多為5. 考慮第i題,選擇四個選項的人數分別為ai, bi, ci, di,所以ai+ bi+ ci+ di = 16. 這題至少會提供 ai 2 + bi 2 + ci 2 + di 2 ≥ 4 4 2 = 24的雙人組對於第i題為相同 答案,所以24m ≤ 162 = 120,也就是m ≤ 5. m = 5的構造:
5. [9 pts] 給定一個圓內接四邊形 ABCD, 令對角線 AC 和 BD 交在 P . 由 P 到 AB 和 CD 的垂足分別為 E 和 F , 而AD 和BC 的中點分別為 M 和N , 證明: EM F N 是一個箏形。
Given a cyclic quadrilateral ABCD. Let P be the intersection of the two diagonals AC and BD. The pedals of the perpendicular lines from P to AB and CD are respectively E and F , and the midpoints of AD and BC are respectively M and N . Prove that EM F N is a deltoid (kite shape).
解: 令P A中點為X,P D中點為Y ,則有XE = P A2 = Y M .同理有Y F = XM . 又因為 XM k P D 且 Y M k P A, 可計算出 ∠MXE = ∠AP D + 180◦ −
2∠BAC = ∠MY F ;故M XE 與F Y M 全等,而得到M E = M F ,同理N E = N F ,故得證。
6. [10 pts]令a1, a2, . . .為正實數數列,而且對任意n ≥ 1有下列不等式 a2n+1 ≥ a 2 1 13 + a2 2 23 + · · · + a2 n n3. 試證:存在一個正整數M 使得 M X n=1 an+1 a1+ a2+ · · · + an > 2016 1016.
Let a1, a2, . . . be a sequence of positive real numbers such that for each n ≥ 1,
a2n+1 ≥ a 2 1 13 + a2 2 23 + · · · + a2 n n3.
Show that there is a positive integer M such that
M X n=1 an+1 a1+ a2+ · · · + an > 2016 1016.
解: From Cauchy-Schwarz inequality, we have
n X k=1 a2 k k3 ! n X k=1 k3 ! ≥ n X k=1 ak !2 . Since a2 n+1 ≥ Pn k=1 a2k k3 and Pn k=1k3 = n(n+1) 2 2 , we get an+1 a1+ a2+ · · · + an ≥ 2 n(n + 1). Therefore, M X n=1 an+1 a1+ a2+ · · · + an ≥ M X n=1 2 n(n + 1) = 2 M X n=1 1 n − 1 n + 1 = 2 − 2 M + 1 ≥ 2016 1016, if M ≥ 126.
