Article 40
The 120o, 45o, 15o Triangle and its Euler Line
Christopher J Bradley
C
30
30
15
15 15
O
75
30
15
P
105
60
S
60
U
Q
90
T
60
R
120
30
A
120
135
45
30
30
15
15
Figure
H
1
B
1. Introduction
We find that the triangle with angles A, B, C equal to 120o, 45o, 15o respectively together with its
Euler line OH, where O is the circumcentre and H the orthocentre, exhibits a number of
interesting results. We define P, Q, R to be the points OH^BC, OH^CA, OH^AB respectively.
We prove in what follows the following results:
(i) PQ = QR;
(ii) OQ = RH;
(iii)Triangle AQR is equilateral;
(iv) If U is the fourth vertex of parallelogram RAQU, then U lies on the circumcircle of ABC;
(v) Circle PAC passes through H;
(vi) Circle CBH passes through O;
(vii)Circle COQ passes through U;
(viii)Circle COQ meets BC again at a point S such that CS = 2SB;
(ix)Circle OBR passes through U;
(x) Circle APR passes through U and S.
As a consequence of these results and angle-chasing angles between various lines are as shown
in the figure above. In the following areal-co-ordinates are used with u, v, w instead of a, b, c,
where
u = 1//(b2 + c2 – a2), v = 1/(c2 + a2 – b2), w = 1/(a2 + b2 – c2)
(1.1)
and ABC as triangle of reference.
2. Values of a, b, c and the co-ordinates of H and O
The sides of ABC are proportional to the Sines of 120o, 45o, 15o so we set a = ½√3, b = ½√2 and
c = ¼(√6 – √2) Then, from Equations (1.1) we find, after normalization so that u + v + w = 1,
that
u = 2 + √3, v = – 1 – (2√3)/3, w = – (√3)/3.
(2.1)
These are now the normalized co-ordinates of the orthocentre, H.
The co-ordinates of the circumcentre O are now ½(v + w, w + u, u + v) = (x, y, z) so that
x = –½(1 + √3), y = 1 + (1/3)√3, z = (1/6)(3 + √3).
(2.2)
3. The Euler line and P, Q, R
The equation of the Euler line through O, H is known to be
(v – w)x + (w – u)y + (u – v)z = 0,
which in this case is
2
(3.1)
(1 + √3)x + (4 + 2√3)y – (5 + 3√3)z = 0.
(3.2)
The co-ordinates of P, Q, R, where the Euler line meets BC, CA, AB respectively are
P(0, (1/3)√3, 1 – (1/3)√3), Q((1/6)(3 + √3), 0, (1/6)(3 – √3)), R(1 + (1/3)√3, – (1/3)√3, 0). (3.3)
4. Results (i) – (iv) and the point U
From the co-ordinates of P, Q, R it may be seen that P + R = 2Q, so that PQ = QR. Similarly O +
H = R + Q, so that OQ = RH.
From the co-ordinates of A, Q, R we find AQ = AR = (1/12)(3√2 – √6) so that triangle AQR is
equilateral. Since U is defined as the fourth vertex of the parallelogram RAQU, its co-ordinates
are given by U = Q + R – A. that is
U((1/2)(1 + √3), – (1/3)√3, (1/6)(3 – √3)).
(4.1)
It may now be checked that U lies on the circumcircle of triangle ABC, whose equation is
3(1 + √3)yz + 2(1 + √3)zx + (√3 – 1)xy = 0.
(4.2)
5. Results (v) – (viii) and the point S
Circle PAC has equation
2√3y2 – (3 + √3)yz – 2(1 + √3)zx + (1 + √3)xy = 0.
(5.1)
It may now be checked that H lies on this circle.
Circle CBH has equation
2x2 + 3(1 + √3)yz + 2(2 + √3)zx + (1 + √3)xy = 0.
(5.2)
It may now be checked that O lies on this circle.
Circle COQ has equation
2x2 + (3 + √3)y2 – 2(3 + √3)yz – 2(2 + √3)zx + 2(1 + √3)xy = 0.
(5.3)
It may now be checked that U lies on this circle. Also the circle COQ meets BC, x = 0, at the
point S with co-ordinates (0, 2/3, 1/3) so that CS = 2BS, a rather unexpected trisection of the side
BC.
6. Results (ix) and (x)
Circle BOR has equation
3
(√3 – 1)x2 – (3 + √3)z2 + 2(3 + √3)yz + 2(1 + √3)zx + 2xy = 0.
(6.1)
It may now be checked that U lies on this circle.
Circle APR has equation
2y2 + 2(1 + √3)z2 – (5 + √3)yz – 4zx + (√3 – 1)xy = 0.
It may now be checked that both U and S lie on this circle.
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4
(6.2)