- IK+ MB M I= IK (dpcm) tr
ld tam gidc Heron
Neu goi a, b, r: le dO dai ciic canh cita tam gi6c Heron theo thrl tu tAng ddn thi a = b - l,
t: = b + 1, nta chu vi vd binh phuong diOn tfch
cfra tam gi6c d6ldn luot ld p =* "a
2
.s, = 3f q)'Ifq)'
-
'].
(2/ [\2i )
Ndu b lh sd le thi 52 khong thd le sd nguy0n. Do d6 b phiri ld sd chin. Dat h = 2k (vdi k lit
sd nguy6n duong) thi 2 = 3k vi,9 = 3E(P - l).
Vi S lA sd nguy6n duong vi 52 i3 nOn ^S2 i9, suy ra kz - | = 3f v6i r lb sd nguyOn duong.
Nhu vAy S = 3kr, trong d6 t ld nfta canh ldn thri'hai vd r li b6n kfnh duong trdn n6i tidp
cira tam gi6c Heron.
Phuong trinh F - I = 3f ld phtong trinh Pell
loai l.
Phuong trinh ndy c6 vO so nghiOm nguy6n duong 1k,,: r,,). Tdt ci cdc nghiOm d6 duoc cho
boi c6c cOng thrlc sau:
, (2* ".6)" + 12 - Jl)"An - 2 ' An - 2 ' (z+Ji),,-(2-Ji)" trt- ^ l; zN)
Hai ddy sd (k,,) vh (r,,) cdn duoc xi{c dinh b6i ciic h0 thftc truy hdi sau:
kt = 2, kz = 7, k,,*, = 4k,,*1- k,,,
/'t = 1, fz= 4, rr*r= 4rr*t- rr'
2. Canh cira tam gi6c Heron
Dg dai cdc canh
"t o tu- gi6c Heron du-o. c tfnh theo c6c cOng thtlc sau:
an = (2*.6), +12-J11" -1,
b, = (2* ..6)' + (2 -.6)',
c,, = (2* 16)' + (2 -',,6)' * l.
Ba ddy sd (a,,), (b,,), (c,) cdn duo.c x6c dinh b&i c6c h0 thrlc truy h6i sau:
at = 3, az= 13, an+z= 4au*t- a,,+ 2,
br = 4, hz= 14, b,*z= 4b,,*t- b,,,
ct = 5, cz= 13, c,,*y= 4c,,*y- c,,-
3. Di6n tich cfia tam gi6c Heron