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Mathematical Olympiad Question
1. For any 3-digit integer,let s be the sum of its digits , r be the sum of the reciprocals of its digits , and p be the product of its digits.Find all 3-digit integer such that s=rp.2. Inside a 10*10 square, at most how many squares of different sizes,with integral side lengths and sides parallel to the large... 顯示更多 1. For any 3-digit integer,let s be the sum of its digits , r be the sum of the reciprocals of its digits , and p be the product of its digits.Find all 3-digit integer such that s=rp.2. Inside a 10*10 square, at most how many squares of different sizes,with integral side lengths and sides parallel to the large square, can be drawn, such that no two small square contain a common interior point? (It is possible that two small square have a common point onj their sides.)3. Given any triangle ABC. For any points D,E,F lying on sides BC,CA,AB respectively , prove that the circumcircles(the unique circle passing through the three points of a triangle) of △ AFE,△BDF,△CED intersect at a single point.
最佳解答:
1)Let 100a + 10b + c be the 3-digit integer , thens = a + b + c r = 1/a + 1/b + 1/c .......(a , b , c ≠ 0) p = abcs = rp a + b + c = (1/a + 1/b + 1/c) abc a + b + c - (ab + bc + ca) = 0 a + b + c - (ab + bc + ca) + abc - 1 = abc - 1 (a - 1) (b - 1) (c - 1) = abc - 1 ∵ (a - 1) (b - 1) (c - 1) ≤ a(b - 1)(c - 1) ≤ ab(c - 1) < abc i.e. abc - 1 ≤ a(b - 1)(c - 1) ≤ ab(c - 1) < abc ∴ abc - 1 = a(b - 1)(c - 1) = ab(c - 1) < abc for only 2 integral values for them. a(b - 1)(c - 1) = ab(c - 1) a(c - 1) = 0 c = 1 since a ≠ 0 , Similarly , a = b = 1 ,The required number = 111. 2)The number of squares < 7 since 12 + 22 + 32 + 42 + 52 + 62 + 72 = 140 > 102For 6 squares , i.e. 12 + 22 + 32 + 42 + 52 + 62 , the sum of the sides of the first 2 largest square = 5 + 6 > 10 , so at most 5 squares , for example : ■ □ □ □ □ ■ ■ ■ ■ ■ □ □ □ □ □ ■ ■ ■ ■ ■ ■ ■ □ □ □ ■ ■ ■ ■ ■ ■ ■ □ □ □ ■ ■ ■ ■ ■ □ □ □ □ □ ■ ■ ■ ■ ■ □ □ □ □ □ □ □ □ □ □ ■ ■ ■ □ □ □ ■ ■ ■ ■ ■ ■ ■ □ □ □ ■ ■ ■ ■ ■ ■ ■ □ □ □ ■ ■ ■ ■ □ □ □ □ □ □ ■ ■ ■ ■ 3) Let H be the another intersection point of the circumcircles of △BDF and △CED. 圖片參考：http://imgcld.yimg.com/8/n/HA04628698/o/701205210027313873407760.jpg ∠ECD = ∠EHG (ext. ∠cyclic quad.) ∠FBD = ∠FHG (ext. ∠s, cyclic quad.) ∴ ∠FHE = ∠EHG +∠FHG = ∠ECD + ∠FBD = 180° - ∠FAE Therefore AFHE is a cyclic quad. (opp. ∠s supp.) The circumcircle of △AFE passing through the point H. ∴ The circumcircles of △ AFE,△BDF,△CED intersect at a single point H.
其他解答:
For Q3, also consider point H outside the triangle...
Mathematical Olympiad Question
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發問:1. For any 3-digit integer,let s be the sum of its digits , r be the sum of the reciprocals of its digits , and p be the product of its digits.Find all 3-digit integer such that s=rp.2. Inside a 10*10 square, at most how many squares of different sizes,with integral side lengths and sides parallel to the large... 顯示更多 1. For any 3-digit integer,let s be the sum of its digits , r be the sum of the reciprocals of its digits , and p be the product of its digits.Find all 3-digit integer such that s=rp.2. Inside a 10*10 square, at most how many squares of different sizes,with integral side lengths and sides parallel to the large square, can be drawn, such that no two small square contain a common interior point? (It is possible that two small square have a common point onj their sides.)3. Given any triangle ABC. For any points D,E,F lying on sides BC,CA,AB respectively , prove that the circumcircles(the unique circle passing through the three points of a triangle) of △ AFE,△BDF,△CED intersect at a single point.
最佳解答:
1)Let 100a + 10b + c be the 3-digit integer , thens = a + b + c r = 1/a + 1/b + 1/c .......(a , b , c ≠ 0) p = abcs = rp a + b + c = (1/a + 1/b + 1/c) abc a + b + c - (ab + bc + ca) = 0 a + b + c - (ab + bc + ca) + abc - 1 = abc - 1 (a - 1) (b - 1) (c - 1) = abc - 1 ∵ (a - 1) (b - 1) (c - 1) ≤ a(b - 1)(c - 1) ≤ ab(c - 1) < abc i.e. abc - 1 ≤ a(b - 1)(c - 1) ≤ ab(c - 1) < abc ∴ abc - 1 = a(b - 1)(c - 1) = ab(c - 1) < abc for only 2 integral values for them. a(b - 1)(c - 1) = ab(c - 1) a(c - 1) = 0 c = 1 since a ≠ 0 , Similarly , a = b = 1 ,The required number = 111. 2)The number of squares < 7 since 12 + 22 + 32 + 42 + 52 + 62 + 72 = 140 > 102For 6 squares , i.e. 12 + 22 + 32 + 42 + 52 + 62 , the sum of the sides of the first 2 largest square = 5 + 6 > 10 , so at most 5 squares , for example : ■ □ □ □ □ ■ ■ ■ ■ ■ □ □ □ □ □ ■ ■ ■ ■ ■ ■ ■ □ □ □ ■ ■ ■ ■ ■ ■ ■ □ □ □ ■ ■ ■ ■ ■ □ □ □ □ □ ■ ■ ■ ■ ■ □ □ □ □ □ □ □ □ □ □ ■ ■ ■ □ □ □ ■ ■ ■ ■ ■ ■ ■ □ □ □ ■ ■ ■ ■ ■ ■ ■ □ □ □ ■ ■ ■ ■ □ □ □ □ □ □ ■ ■ ■ ■ 3) Let H be the another intersection point of the circumcircles of △BDF and △CED. 圖片參考：http://imgcld.yimg.com/8/n/HA04628698/o/701205210027313873407760.jpg ∠ECD = ∠EHG (ext. ∠cyclic quad.) ∠FBD = ∠FHG (ext. ∠s, cyclic quad.) ∴ ∠FHE = ∠EHG +∠FHG = ∠ECD + ∠FBD = 180° - ∠FAE Therefore AFHE is a cyclic quad. (opp. ∠s supp.) The circumcircle of △AFE passing through the point H. ∴ The circumcircles of △ AFE,△BDF,△CED intersect at a single point H.
其他解答:
For Q3, also consider point H outside the triangle...
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