A210643 -id:A210643 - cp's OEIS frontend

A182171 Area A of the triangles such that A, the sides and three perpendicular bisectors are integers.

108, 384, 432, 768, 972, 1536, 1728, 2700, 3072, 3456, 3888, 5292, 6144, 6912, 8748, 9600, 10800, 12288, 13068, 13824, 15552, 17280, 18252, 18816, 18900, 19200, 21168, 24300, 24576, 27000, 27648, 31104, 31212, 34560, 34992, 37632

Offset: 1

Michel Lagneau, Apr 16 2012

nonn

Let a triangle with the angles (A, B, C) and the sides opposite the angles labeled (a, b, c). The length of the perpendicular bisectors is given by (x, y, z) where:
x is the perpendicular bisector passing through the midpoint of the segment BC = a;
y is the perpendicular bisector passing through the midpoint of the segment AC = b;
z is the perpendicular bisector passing through the midpoint of the segment AB = c.
We obtain the relations:
x = (a/2)*tg B if x intersects AB or (a/2)* tg C if x intersects AC;
y = (b/2)* tg A if y intersects AB or (b/2)* tg C if y intersects BC;
z = (c/2)*tg A if z intersects AC or (c/2) *tg B if z intersects BC.
The area A of the triangle whose sides have lengths a, b, and c is given by Heron's formula: A = sqrt(s*(s-a)*(s-b)*(s-c)), where s = (a+b+c)/2.
Finally, we obtain:
x = (a/2) * min {tg B, tg C }; y = (b/2) * min {tg A, tg C }; z = (c/2) * min {tg A, tg B } with tg A = 4*A/(b^2+c^2-a^2) ; tg B = 4*A/(c^2+a^2-b^2) ; tg C = 4*A/(a^2+b^2-c^2).
Properties of this sequence:
The numbers of the form 108*n^2, 384*n^2, 768*n^2, 17280*n^2, 18900*n^2 are in the sequence because the area of the primitive triangles (15, 15, 18), (24, 32, 40), (40, 40, 64), (120, 288, 312), (150, 255, 315) are 108, 384, 768 , 17280 and 18900 respectively.
There exists three class of numbers included into a(n) :
Case (i) : a subset of isosceles triangles;
Case (ii) : a subset of right triangles;
Case (iii) : other (neither isosceles nor right triangle).

			Primitive solutions follow:
Area,  ( a,   b,   c),  ( x,   y,   z), Case
  108,  (15,  15,  18),  (10,  10,  12), Isosceles,
  384,  (24,  32,  40),  (16,  12,  15), Right,
  768,  (40,  40,  64),  (15,  15,  24), Isosceles,
17280, (120, 288, 312), (144,  60,  65), Right,
18900, (150, 255, 315), (100,  68,  84), Other,
27000, (255, 255, 450),  (68,  68, 120), Isosceles,
34560, (312, 312, 576),  (65,  65, 120), Isosceles,
53760, (272, 400, 448), (255, 150, 168), Other,
54000, (240, 450, 510), (225, 120, 136), Right,
91476, (429, 462, 495), (364, 308, 330), Other,
95256, (252, 819, 945), (168, 104, 120), Other,
96768, (336, 720, 960), (126, 105, 140), Other.

Ray Chandler, Table of n, a(n) for n = 1..65
Eric W. Weisstein, MathWorld: Triangle

Cf. A188158, A210643, A181924, A181925, A181928.

zz:=evalf(1/10^6):k:=0:nn:=350:
for a from 15 to nn do:
   for b from a to nn do:
       for c from b to nn do:
             p:=(a+b+c)/2:s:=p*(p-a)*(p-b)*(p-c):
             u:= a^2+b^2-c^2:v:= b^2+c^2-a^2 :w:=c^2+a^2-b^2:
             if s>0 then s1:=sqrt(s):else fi:
             if u>0 then u0:=u:else u0:=zz:fi:
             if v>0 then v0:=v:else vo:=zz:fi:
             if w>0 then w0:=w:else w0:=zz:fi:
a0:= evalf(2*a*s1/w0):a1:=evalf(2*a*s1/u0): b0:= evalf(2*b*s1/u0):b1:=evalf(2*b*s1/v0): c0:= evalf(2*c*s1/w0):c1:=evalf(2*c*s1/v0):
             if a0

x = Min{2*a*A/(c^2+a^2-b^2) , 2*a*A/(a^2+b^2-c^2)};
y = Min{2*b*A/(a^2+b^2-c^2)  , 2*b*A/(b^2+c^2-a^2)};
z = Min{2*c*A/(c^2+a^2-b^2)  , 2*c*A/(b^2+c^2-a^2)}.

More terms from Ray Chandler, Apr 24 2013

A210645 Area A of the triangles such that A, the sides and one of the altitudes are four consecutive integers of an arithmetic progression d.

Original entry on oeis.org

84, 336, 756, 1344, 2100, 3024, 4116, 5376, 6804, 8400, 10164, 12096, 14196, 16464, 18900, 21504, 24276, 27216, 30324, 33600, 37044, 40656, 44436, 48384, 52500, 56784, 61236, 65856, 70644, 75600, 80724, 86016, 91476, 97104, 102900, 108864, 114996, 121296

Offset: 1

Michel Lagneau, Mar 26 2012

nonn

See the first link for the comments. We use Heron's Formula for a triangle: A = sqrt(s*(s-a)*(s-b)*(s-c)), where s = (a+b+c)/2.The sides and one of the altitudes are of the form q+d, q, q-d, q-2d.

			84 is in the sequence because (a, b, c, h) = (15, 14, 13, 12) => A = sqrt(21*(21-15)*(21-14)*(21-13)) = sqrt(7056) = 84 but A = (1/2)*h*b = (1/2)*12*14 = 84.

Eric Weisstein, Altitude

Cf. A188158, A210643.

with(numtheory):T:=array(1..1000):k:=0:nn:=800:for q from 1 to nn do: for d from 1 to nn do: a:=q+d:b:=q:c:=q-d:h1:=q-2*d:p:=(a+b+c)/2:s:=p*(p-a)*(p-b)*(p-c):if s>0 then s1:=sqrt(s): h11:=2*s1/a: h22:=2*s1/b:h33:=2*s1/c:if s1=floor(s1) and (h1=h11 or h1=h22 or h1=h33)  then k:=k+1:T[k]:=s1:else fi:fi:od:od: L := [seq(T[i],i=1..k)]:L1:=convert(T,set):A:=sort(L1, `<`): print(A):

Conjecture: a(n) = 84*n^2. a(n) = 3*a(n-1)-3*a(n-2)+a(n-3). G.f.: 84*x*(1+x)/(1-x)^3. - Colin Barker, Apr 19 2012

A230558 Integer areas of extouch triangles of integer-sided triangles.

Original entry on oeis.org

30, 48, 72, 84, 120, 192, 252, 270, 288, 336, 432, 480, 648, 720, 750, 756, 768, 780, 936, 1008, 1080, 1152, 1200, 1344, 1470, 1728, 1800, 1920, 2100, 2268, 2352, 2400, 2430, 2592, 2784, 2880, 3000, 3024, 3060, 3072, 3120, 3528, 3600, 3630, 3888, 4032, 4116

Offset: 1

Michel Lagneau, Oct 23 2013

nonn

The extouch triangle T1T2T3 is the triangle formed by the points of tangency of a triangle ABC with its excircles J1, J2 and J3. The points T1, T2, and T3 can also be constructed as the points that bisect the perimeter of the triangle ABC starting at A, B, and C.
The side lengths of the extouch triangle are:
a'= sqrt(a^2 - 4*A^2/b*c)
b'= sqrt(b^2 - 4*A^2/a*c)
c'= sqrt(c^2 - 4*A^2/a*b)
where A is the triangle area of the original triangle.
The extouch triangle has area:
A*(a+b-c)*(a-b+c)*(-a+b+c)/4abc = A*2*r^2*s/(a*b*c) where r and s are the inradius and semiperimeter, respectively.
It is interesting to note that the sides of the extouch triangles are irrational numbers (in the general case) but the areas are integers.
The following table gives the first values (A', A, a, b, c,t1,t2,t3) where A' is the area of the extouch triangles, A is the area of the triangles ABC, a, b, c the integer sides of the original triangles ABC and t1, t2, t3 are the integer sides of the extouch triangles.
-------------------------------------------------------------
A'  |   A  | a | b  | c  |     t1    |     t2    |  t3
-------------------------------------------------------------
30  |  150 | 15| 20 | 25 | 3*sqrt(5) | 4*sqrt(10)|5*sqrt(13)
48  |  300 | 25| 25 | 40 | sqrt(265) | sqrt(265) | 32
72  |  300 | 25| 25 | 30 | sqrt(145) | sqrt(145) | 18
84  | 1050 | 35| 75 |100 | 7*sqrt(13)|3*sqrt(385)|8*sqrt(130)
120 |  600 | 30| 40 | 50 | 6*sqrt(5) |8*sqrt(10) |10*sqrt(13)
192 | 1200 | 50| 50 | 80 |2*sqrt(265)|2*sqrt(265)| 64
252 | 2100 | 35|120 |125 |    7      |72*sqrt(2) |5*sqrt(457)
270 | 1350 | 45| 60 | 75 |9*sqrt(5)  |12*sqrt(10)|15*sqrt(13)
288 | 1200 | 50| 50 | 60 |2*sqrt(145)|2*sqrt(145)| 36
336 | 4200 | 70|150 |200 |14*sqrt(13)|6*sqrt(485)|16*sqrt(130)
432 | 2700 | 75| 75 |120 |3*sqrt(265)|3*sqrt(265)| 96
480 | 2400 | 60| 80 |100 |12*sqrt(5) |16*sqrt(10)|20*sqrt(13)
648 | 2700 | 75| 75 | 90 |3*sqrt(145)|3*sqrt(145)| 54
..................................................
Observation: the three altitudes of a majority of initial triangles ABC are integers, except very rare triangles, for example the initial triangle (35, 120, 125) where A = 2100 (see the following table).
This table gives the first values (A',A, h1, h2, h3) where A' is the area of the extouch triangles, A is the area of the initial triangles ABC and h1, h2, h3 are the altitudes of the initial triangles.
-------------------------------
A'  |  A   |  h1 |  h2 |  h3
-------------------------------
30  |  150 |  20 |  15 |  12
48  |  300 |  24 |  24 |  15
72  |  300 |  24 |  24 |  20
84  | 1050 |  60 |  28 |  21
120 |  600 |  40 |  30 |  24
192 | 1200 |  48 |  48 |  30
252 | 2100 | 120 |  35 | 168/5
270 | 1350 |  60 |  45 |  36
288 | 1200 |  48 |  48 |  40
336 | 4200 | 120 |  56 |  42
432 | 2700 |  72 |  72 |  45
480 | 2400 |  80 |  60 |  48
648 | 2700 |  72 |  72 |  60
...............................

			30 is in the sequence. We use two ways:
First way: the formula A' = A*(a+b-c)*(a-b+c)*(-a+b+c)/(4*a*b*c) gives directly the result: A' = 150*(15+20-25)*(15-20+25)*(-15+20+25)/(4*15*20*25) = 30, with the area A = 150 obtained by Heron's formula A = sqrt(s*(s-a)*(s-b)*(s-c)) = sqrt(30*(30-15)*(30-20)*(30-25)) = 150, where s is the semiperimeter.
Second way: by calculation of the sides t1, t2, t3 and by using Heron's formula.
The extouch triangle (t1,t2,t3) of the initial triangle (a, b, c) = (15, 20, 25) is the triangle (3*sqrt(5), 4*sqrt(10), 5*sqrt(13)) where:
a' = sqrt(a^2 - 4*A^2/b*c) = sqrt(15^2-4*150^2/(20*25)) = 3*sqrt(5);
b' = sqrt(b^2 - 4*A^2/a*c) = sqrt(20^2-4*150^2/(15*25)) = 4*sqrt(10);
c' = sqrt(c^2 - 4*A^2/a*b) = sqrt(25^2 - 4*150^2/(15*20)) = 5*sqrt(13).
Now, we use Heron's formula with (t1,t2,t3). We find A'=sqrt(s1*(s1-t1)*(s1-t2)*(s1-t3))with:
s1 =(t1+t2+t3)/2 = (3*sqrt(5)+ 4*sqrt(10) + 5*sqrt(13))/2;
We find A'= 30.

T. Dosa, Some Triangle Centers Associated with the Excircles, Forum Geometricorum, Volume 7 (2007) 151-158.
C. Kimberling, Triangle Centers and Central Triangles, Congr. Numer. 129, 1-295, 1998.
Eric Weisstein's World of Mathematics, Extouch Triangles

Cf. A188158, A210643.

nn = 1000; lst = {}; Do[s = (a + b + c)/2; If[IntegerQ[s], area2 = s (s - a) (s - b) (s - c); t = Sqrt[area2]*(a + b - c)*(a - b + c)*(-a + b + c)/(4*a*b*c); If[0 < area2 && IntegerQ[Sqrt[area2]] && IntegerQ[t], AppendTo[lst, t]]], {a, nn}, {b, a}, {c, b}]; Union[lst]

A256579 Integer areas of integer-sided triangles where at least one of the three altitudes is of prime length.

Original entry on oeis.org

6, 12, 30, 60, 84, 168, 330, 546, 660, 1092, 1224, 1710, 2448, 3036, 3420, 6072, 6090, 7440, 12180, 12654, 14880, 17220, 19866, 25308, 25944, 34440, 37206, 39732, 51330, 51888, 56730, 74412, 75174, 89460, 97236, 102660, 113460, 123240, 142926, 150348, 176220

Offset: 1

Michel Lagneau, Apr 02 2015

nonn

Subset of A226453.
The corresponding primes are: 3, 3, 5, 5, 7, 7, 11, 13, 11, 13, 17, 19, 17, 23, 19, 23, 29, 31, 29, 37, 31, 41, 43, 37, 47, 41, 53, 43, 59, 47, 61, 53, 67, 71, 73, 59, 61, 79, 83, 67, 89, ...
The area A of a triangle whose sides have lengths a, b, and c is given by Heron's formula: A = sqrt(s*(s-a)*(s-b)*(s-c)), where s = (a+b+c)/2.
The altitudes of a triangle with sides length a, b, c and area A have length given by Ha = 2*A/a, Hb = 2*A/b, Hc = 2*A/c.
Properties of this sequence:
- The sequence is infinite (see the formula below);
- The prime altitude of a triangle is the greatest prime divisor of a(n) (the proof is easy if we observe the formula);
- There exists two subsets of numbers included into a(n):
Case (i): A subset with right triangles (a,b,c) where a^2 + b^2 = c^2 with area a1(n) = {6, 30, 84, 330, 546, 1224, ...}. The lengths of the prime altitudes are Ha or Hb = a = p. The sides are of the form (p, q, q+1) with p = sqrt(2*q+1) => the sides are equal to (p, (p^2-1)/2, (p^2+1)/2) and a(n) = (p^3-p)/4.
Case (ii): A subset with isosceles triangles formed by two right triangles of the sequence. So, the areas are a2(n) = {12, 60, 168, 660, 1092, 2448, ...} = 2*a1(n). The sides are of the form (a, a, 2*(a-1)) = ((p^2+1)/2, (p^2+1)/2, p^2-1) and Ha = sqrt(2*a-1) = p, a2(n) = 2*a1(n) = (p^3-p)/2.
We did not find a class of non-isosceles and non-right triangles (a, b, c) whose three altitudes include one of prime length.
The following table gives the first values (A, a, b, c, Ha, Hb, Hc) where A is the integer area, a, b, c are the sides and Ha <= Hb <= Hc are the altitudes.
+------+-----+-----+-----+----------+----------+---------+
|   A  |  a  |  b  |  c  |    Ha    |   Hb     |   Hc    |
+------+-----+-----+-----+----------+----------+---------+
|    6 |   3 |   4 |   5 |   12/5   |    3     |    4    |
|   12 |   5 |   5 |   8 |    3     |   24/5   |   24/5  |
|   30 |   5 |  12 |  13 |    5     |   60/13  |   12    |
|   60 |  13 |  13 |  24 |    5     |  120/13  |  120/13 |
|   84 |   7 |  24 |  25 |  168/25  |    7     |   24    |
|  168 |  25 |  25 |  48 |    7     |  336/25  |  336/25 |
|  330 |  11 |  60 |  61 |  660/61  |   11     |   60    |
|  546 |  13 |  84 |  85 | 1092/85  |   13     |   84    |
|  660 |  61 |  61 | 120 |   11     | 1320/61  | 1320/61 |
| 1092 |  85 |  85 | 168 |   13     | 2184/85  | 2184/85 |
| 1224 |  17 | 144 | 145 | 2448/145 |   17     |  144    |
| 1710 |  19 | 180 | 181 | 3420/181 |   19     |  180    |
| 2448 | 145 | 145 | 288 | 4896/145 | 4896/145 |   17    |
+------+-----+-----+-----+----------+----------+---------+

Eric Weisstein's World of Mathematics, Altitude
Eric Weisstein's World of Mathematics, Isosceles Triangle
Eric Weisstein's World of Mathematics, Right Triangle

Cf. A188158, A210643, A210645, A226453.

# program using the formula
lst:={}:for n from 2 to 50 do:p:=ithprime(n):p1:=(p^3-p)/4:p2:=(p^3-p)/2:lst:=lst union {p1} union {p2}:od:print(lst):

nn = 300; lst = {}; Do[s = (a + b + c)/2; area2 = s (s - a) (s - b) (s - c); If[area2>0 && IntegerQ[Sqrt[area2]]&&(PrimeQ[(2*Sqrt[area2])/a]|| PrimeQ[(2*Sqrt[area2])/b]||PrimeQ[(2*Sqrt[area2])/c]), AppendTo[lst, Sqrt[area2]]], {a, nn}, {b, a}, {c, b}]; Union[lst]

a(n) = (prime(n)^3 - prime(n))/4 for the right triangles;

a(n) = (prime(n)^3 - prime(n))/2 for the isosceles triangles.

A256584 Integer areas of integer-sided triangles where at least one of the three altitudes is of perfect square length.

Original entry on oeis.org

6, 12, 54, 96, 108, 126, 144, 180, 192, 216, 234, 240, 264, 270, 336, 360, 378, 408, 480, 486, 504, 522, 540, 594, 600, 744, 750, 756, 864, 900, 972, 990, 1008, 1026, 1116, 1134, 1224, 1296, 1350, 1386, 1404, 1494, 1500, 1536, 1584, 1620, 1656, 1728, 1800, 1872

Offset: 1

Michel Lagneau, Apr 02 2015

nonn

a(n) contains A210643.
There are triangles with rational square, for instance, with the area 144, we find for (a,b,c)=(6,50,52) the altitudes {Ha,Hb,Hc} = {72/13, 144/25, 48} but with the same area we find also for (a,b,c)=(18,20,34) the altitudes {Ha,Hb,Hc} = {144/17, 72/5, 16}.
The corresponding squares are 4, 4, 9, 16, 9, 9, 16, 9, 16, 36, 9, 16, 16, 36, 16, 9, 36, 16, 16, 36, 16, 36, 36, 36, 16, 16, 25, 36, 36, 36, 36, 36, 16, 36, 36, 36, 144, 144, 36, 36, 36, 36, 100, 64, 144, 36, 36, 36, 144, 36, ...
The subsequence of the primitive triangles are 6, 12, 126, 144, 180, 216, 234, ...
The area A of a triangle whose sides have lengths a, b, and c is given by Heron's formula: A = sqrt(s*(s-a)*(s-b)*(s-c)), where s = (a+b+c)/2.
The altitudes of a triangle with sides length a, b, c and area A have length given by Ha= 2A/a, Hb= 2A/b, Hc= 2A/c.
Properties of this sequence:
- The sequence is infinite because from de initial primitive triangle (3,4,5), the area A’ of the triangle (3*3^m, 4*3^m, 5*3^m) is also in the sequence where A’ = 6*3^2m and {Ha, Hb, Hc} = {4*3^m, 3^(m+1), (4*3^(m+1))/5}. The altitude Ha or Hb is square.
- There are three subsets of numbers included into a(n):
Case (i): A subset with right triangles (a,b,c) where a^2+b^2 = c^2 with area a2(n) = {6, 54, 96, 180, 240, 270, ...}
Case (ii): A subset with isosceles triangles of area a1(n)= {12, 108, 192, 360, 480, 540, ...} = 2*a1(n).
Case (iii): A subset with non-isosceles and non-right triangles of area a3(n)= {126, 144, 216, 234, 264, 336, ...}
-----------------------------------------------------------
|    A |   a |   b  |   c |    Ha    |    Hb   |   Hc     |
-----------------------------------------------------------
|    6 |   3 |   4  |   5 |   12/5   |    3    |    4     |
|   12 |   5 |   5  |   6 |     4    |   24/5  |   24/5   |
|   54 |   9 |  12  |  15 |   36/5   |    9    |   12     |
|   96 |  12 |  16  |  20 |    48/5  |   12    |   16     |
|  108 |  15 |  15  |  24 |     9    |   72/5  |   72/5   |
|  126 |  15 |  28  |  41 |   252/41 |    9    |   84/5   |
|  144 |  18 |  20  |  34 |  144/17  |   72/5  |   16     |
|  180 |   9 |  40  |  41 |  360/41  |    9    |   40     |
|  192 |  20 |  20  |  24 |    16    |   96/5  |   96/5   |
|  216 |  12 |  39  |  45 |   48/5   |  144/13 |   36     |
|  234 |  15 |  41  |  52 |    9     |  468/41 |  156/5   |
|  240 |  16 |  30  |  34 |  240/17  |   16    |   30     |
|  264 |  33 |  34  |  65 |  528/65  |  264/17 |   16     |

			1350 is in the sequence because the altitudes of the triangle (45, 60, 75) are (60, 45, 36).

Eric Weisstein, Altitude
Eric Weisstein, Isosceles Triangle
Eric Weisstein, Right Triangle

Cf. A210643.

nn=200;lst={};Do[s=(a+b+c)/2;If[IntegerQ[s],area2=s (s-a) (s-b) (s-c);If[0

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A182171 Area A of the triangles such that A, the sides and three perpendicular bisectors are integers.

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A210645 Area A of the triangles such that A, the sides and one of the altitudes are four consecutive integers of an arithmetic progression d.

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A230558 Integer areas of extouch triangles of integer-sided triangles.

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A256579 Integer areas of integer-sided triangles where at least one of the three altitudes is of prime length.

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A256584 Integer areas of integer-sided triangles where at least one of the three altitudes is of perfect square length.

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