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On Triangles with Sides That Form...
УДК 514.112.3
On Triangles with Sides That Form an Arithmetic Progression
Yu.N.Maltsev,A.S.Monastyreva
Altai State Pedagogical University (Barnaul, Russia)
О треугольниках, стороны которых образуют арифметическую прогрессию
Ю.Н.Малъцев, А.С. Монастырева
Алтайский государственный педагогический университет (Барнаул, Россия)
Properties of triangles such that the squares of their sides form an arithmetic progression were studied in 2018. In this paper, triangles with sides that form an arithmetic progrèssion aredescribed.Let a, b, c besidesof an arbitrary triangleABC.Ifsides b,a, c of thetribngkABCform an arithmetic progression then, for examp le, the equality a=(bgg)/2(b<a<c)bglds.The class of triangles for which a=(fgc)i2 iseseaierShaneheclebs fflriangies èor wbich bsa,e form an arithmctisgeggsession.enihingapcr,we study the proptrties of triangles for which this equality holds. Thus, triangles with sides that form an arithmetic progression are described with the help of the parameters pc n^.Claes es oCeectengslar taianglesgsiangles wiih an)Se 3b(, feianglea wiilDngee 60°, lsiengleswiehsai=ie 120° are s^i^c^isa^i^ctb^^^riB^e)^.
KanworOf triinglr,clsaumr=diusiinranius.semibeffmsl ter, arltBmeiicpragsestiac.
DOII 0.14258/izvasu(2020)1-18
В 2018 г. описаны свойства треугольников, у которых квадраты сторон образуют арифметическую прогрессию. В данной работе описываются треугольники, стороны которыхобра зука арифметич ес^ь^ц^ пр огр ес-сию. Пусть a, Ь, c — это стороны произвольного треугольника ABC. Если сгороны1Ь, а, с тре)гооьникз ABо образуют ари+л етическую прогрессию, то, например, следующее раленство выполняется: a=(b+c)/2 (Ь < a < с). Однткокл асг тл еуголышков,слякотс°ыхуавкпс тво гс(р+с)/2выполняезтя,боиьше,ч емкл асс т^^ оикни-бко, сто роны кыторых Ь, а.зобртз^от арифметическую псн гр)аслю.у рассоящейработзмыизупссм (воргстна бр(^ло;^ьни(гидн^отс^]^с^н которого вошолрнатря это равинство. И этн пкбволяетобисать(ноязыкгпахаме-трпс р)ХоС) треугооьникн, сто.сныооторых оброзуют арифметич еслую прогрессию. Отдельно описываются классы треугольников:прямоугольные треуг олоники, треугольники сугломнО°,троурольники с углом 60°, треугольники с углом 120°.
Ключевые слова: треугольник, радиус описанной окружности, радиус вписанной окружности, полупериметр, арифметическая прогрессия.
1. Introduction. Let R and r be the
circumradius and the inradius of an arbitrary triangle ABC. Also, let AB = c,AC = b, BC = a,
ZA = a, ZB = в, ZC = y, p = a + Ь + C.
In [1], the authors studied properties of triangles such that the squares of its sides form an arithmetic progression. Also, description of such triangles associated with its remarkable points is given. It is natural to describe triangles such that its sides form an arithmetic progression. In this case, for examples, b + c
the equality a = —^— (b < a < c) holds. The class
of triangles for which a
b + c .
is greater than the
class of triangles for which b, a, c form an arithmetic progression. Really, the class of triangles for which
a = +C also contains all equilateral triangles. Further, this class contains rectangular triangle with sides 3,4, 5. In [2, Ex. 352], [3, Ex. 286, 287, 321], [4, Ex. 11.55, 12.23], [5, p. 89, Ex. 38], and [6, Ex. 88], some properties and characterizations of such triangles are given.
In this paper, study of this triangles is continued and the following theorems are proved [7].
H3BecTHii AaTry.MaTeMaTHKa u MexaHHKa. 2020. № 1 (111)
Theorem 1. For arbitrary triangle ABC, the following conditions are equivalent: b + c
1. a =
2
2. p2 = 18Rr - 9r2;
3. the sides of the triangle ABC are equal to | - V2r(R - 2r), | + V2r(R - 2r). Besides this, a < 60°.
Theorem 2. Let R,r be arbitrary positive numbers such that R > 2r and p = \/18Rr - 9r2. Then there exists a unique triangle ABC such that R, r, p are the circumradius, the inradius, the semiperimeter of the triangle ABC respectively. Theorem 3. (1) Let A ABC be a rectangular
b + c
triangle. The equality a -
5
2
holds in ABC
14
2
R = — r (i.e. in this case, AABC is homothetic to a triangle with sides 6,10,14).
2. Proof of the main results.
Proof of Theorem 1. Let AABC be an
b + c , 2p
arbitrary triangle such that a = ——. Then a = —.
By [8], a, b, c are roots of the equation
x3 - 2px2 + (p2 + r2 + 4Rr)x - 4pRr = 0. (1)
Therefore
2p
3
2p
3
2p
IT " 2p +(p2 + r2 + 4Rr) — -4pRr = 0
3
and p2 = 18Rr - 9r2. So (1) implies (2).
Now prove that (2) implies (1). Assume that p2 = 2p
18Rr-9r2. Then — is a root of the equation (1) and
x3 - 2px2 + (p2 + r2 + 4Rr)x - 4pRr =
= (x - ^ (X2 - 4px + 6Rr^ .
It implies that the numbers 2p
b = | -y/ 2r(R - 2r), a =
2p
c = 2p + \/2r(R - 2r)
2p b + c
are roots of (1) and the equality a = — = —-— holds in AABC. Note that we have proved also that the condition (1) is equivalent to the condition (3). We will show a = ZBAC < 60°. We need to prove the following lemma.
r
Lemma. cos a =1 ——.
R
b+c
Proof. The equality a = —— is equivalent to
a2 = (b + c — a)2. In its turn, the condition a2 = (b + c - a) 2 is equivalent to the equality
1
b2 + c2 a2
2bc 2 (l-
a2 + c2 - b2
2ac
1
+ b2 - c2 2ab
and hence
iff p = 6r,R = -r (i.e. in this case, AABC is homothetic to a triangle with sides 3,4, 5).
(2) Let AABC be a triangle with angle 60°. The
b+c
equality a = —-— holds in AABC iff AABC is a equilateral triangle.
(3) Let AABC be a triangle with angle 120°. The
b + c
equality a = —-— holds in AABC iff p = 5\/3r,
1 — cos a 1 — cos 8 1 — cos y -^-= 4---—- •----.
The last equality can be written as sin — =
/1-2 P -2 Y • • a 0-^-7 n fol 4sin — sin —, i.e. sin — = 2si^ —si^—. By [8], 2 2 2 2 2 a P 7 r sin — sin — sin — = -p-. Thus 2 2 2 4R
• 2 a o • a • P • 7 r sin — = 2 sin — sin — sin — = ——.
2
2 2 2 2R'
Hence cos a =1 - 2 sin2 — = 1 —— and the lemma 2R
is proved.
It is known that R > 2r (see [8-10]). Therefore r1
cos a = 1 - — > - and a < 60°. The proof is R2
complete.
Proof of Theorem 2. Assume that R, r are arbitrary positive numbers such that R > 2r and p = a/18r— - 9r2. By [8, Theorem 2, P. 54], positive numbers R,r,p are the circumradius, the inradius, the semiperimeter of some triangle respectively iff the condition
(p2 - 2R2 - 10Rr + r2)2 < 4R(R - 2r)3 (2)
holds. Note that p = V18Rr - 9r2 implies (2). Really, we have
(p2 - 2R2 - 10Rr + r2)2 = 4(R- 2r)4 < 4R(R- 2r)3.
The proof is complete.
Proof of Theorem 3. Prove (1). Consider a rectangular triangle AABC in which 7 = 90°. By
[8, P. 26], cos a cos P cos 7 = ^ (p2 - (2R + r)2).
Since 7 = ZACB = 90°, we have p = 2R + r. By
b + c
Theorem 1, the equality a = —— is equivalent to
the condition p2 = 18Rr - 9r2 = (2R + r)2, or 2R2 -7Rr + 5r2 = (R - r)(2R - 5r) = 0. Since R > 2r
2
a
3
2
On TriangleswithSides Tha^orm.
b + c
(the Euler's inequality), the equality a = —-— is
equivalent to R = ^ for the rectangular triangle
AABC. So p = 2R + r = 6r, a = y = 4r, c = 2R = 5r, b = 3r for such triangle. Thus AABC is homothetic to a triangle with sides 3,4,5.
Further prove (2). Let AABC be a triangle with angle 60°. By [8], the numbers cos a, cos 3, cos y are roots of the equation
4R2x3 - 4R(R + r)x2 + (p2 + r2 - 4R2)x+
+ (2R + r)2 - p2 =0. (3)
1
b + c .
V3
2
c =2P W2r(R - 2r) = ±4= r
and AABC is homothetic to a triangle with sides 6,10,14.
The proof is complete.
Let ABC be a triangle with angle 30°. Then cos 30° is a root of (3), i.e.
4R2 . ^ - 4R(R + r)3 + (p2 + r2 - 4R2) +
8
V3 2
+ (2R + r)2 - p2
R + r(2 + V3) = p.
b+c
If a = —-—, then, by theorem 1, p2 = 18Rr - 9r2,
i.e. 18Rr- 9r2 = R2 +2(2 + V3)Rr + r2(7 + 4^3), or R2 - 2Rr(7 - V3) + r2(16 + 4^3) = 0. Hence
either R = (4 + 2±3)r, or R = (10 - 4±3)r, and p = R + r(2 + i/3). Consider these cases. Case 1. Let R = (4 + 2±3)r. In this case, by Theorem 1, the sides of the triangle ABC are equal to
a =2p = =(R + r(2 + ±3)) = 2
= 3(6 + 3^ = (4 + 2 V^r,
b = 23F 2r(R - 2r) =
Hence cos 60° = - is a root of (3) and p = \/3(R+r).
b+c
By Theorem 1, the equality a = —— is equivalent
to the condition p2 = 18Rr — 9r2. Since p2 = 3(R2 + 2Rr + r2), we have 3R2 + 6Rr + 3r2 = 18Rr — 9r2,
b+c
i.e. (R — 2r)2 = 0. Therefore the equality a = —-—
is equivalent to R = 2r for any triangle with angle 60°. By [8, P. 8], the last inequality holds if and only if AABC is an equilateral triangle.
Finally, prove (3). Let AABC be a triangle with
angle 120°. Then cos 120° = — 1 is a root of (3) and 3R + r 2
p = -. Thus, by Theorem 1, the equality a =
3(R + r(2 + V3)) 2r(2 + 2V3)r : = (^4 + 2^3- 2\JV3+1
:= 2p+V2r(R - 2r) = (4 + 2^+2^+^ r
and AABC is homothetic to a triangle with sides
4 + 2V3 + 2\!V3+ ^ , 4 + 2^3,
(^4 + 2^3 - 2\!V3+1
is equivalent to the condition p2 = 18Rr -
2 9R2 + 6Rr + r2 14 9r2 =---or R = — r. The last equality
implies
3R + r r ^ 2p 10V3
p = = 5^3r, a = y = _r,
b = y - i/2r(R - 2r) = 2V3r, 2p
Case 2. Let R =(10 — 4±3)r. In this case, the sides and the semiperimeter of the triangle ABC are equal to
2® 2
a = -3 = 2 (12 — 3V3 = (8 — 2\/3)r, p = R + r(2 + ±3) = (12 — 3±3)r,
b = y -a/2r(R - 2r) = (10 - 4\/3)r, c = 6r.
Then AABC is homothetic to a triangle with sides
8 - 2^3,10 - 4v3 6.
So the following theorem is true.
Theorem 4. Let ABC be a triangle with angle b+c
30° and the equality a = —2— holds in AABC. Then either AABC is homothetic to a triangle with sides 8 - 2v310 - 4V3, 6, or AABC is homothetic to a triangle with sides
4 + 2^3 + 2\!V3+ ^ , 4 + 2^3,
^4 + 2^3 - 2\!V3 + 1
Известия АлтГУ. Математика и механика. 2020. № 1 (111)
Consider the class K(p) of triangles with fixed
b + c
angle p (0 < p < n) such that the equality a = —-—
holds for sides a, b, c (in particular, if b, a, c form an arithmetic progression). By [8, p. 26],
p2(1 - cos p) = 4R2 cos3 p - 4R(R + r)cos2 p+ + (r2 - 4R2) cos p + (2R + r)2 = = R2 (4 cos3 p - 4 cos2 p - 4 cos p + 4)+ + Rr ■ 4(1 - cos2 p) + r2(1 + cos p), 2
each Xi (i = 1,2), we can calculate Ri = Xir, pi = Vl8Rir - 9r2 = rV18Ai - 9, ai = 2Pi, bi =
p2 = R24sin2 p+4(1+cosp)Rr+ctg2pr2 = 18Rr-9r2. R
Let A = — > 2. Then A is a root of
r
4 sin2 p ■ t2 + (4 cos p - 14)t + (ctg2 p + 9) = 0. (4)
The equation (4) has no more than two different roots A1,A2. Let r be a fixed positive number. For
3
- V2r(Ri - 2r) ,a = + V2r(Ri - 2r ) ,i < 2. So the class K(p) has no more than two different subclasses of homothetic triangles. For example, we have:
1. the class K(90°) consists of triangles that are homothetic to a triangle with sides 3,4, 5;
2. the class K(30°) consists of triangles that are homothetic to a triangle with sides either 8 — 2\/3, 6, 10 —
4^3, or (4 + 2v3 + 2Vv3+ l) , 4 +
2^3, (4 + 2^3 - 2VV3 + l) ;
3. the class K(60°) consists of equilateral triangles;
4. the class K(120°) consists of triangles that are homothetic to a triangle with sides 3,5,7.
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