Mathematical Olympiad Question

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.

For Q3, also consider point H outside the triangle...