SQUARE AND EQUILATERAL TRIANGLES

A Corollary of SQUARE PROBLEM

Let ABCD be a square, E,E' two points on BC such that DEE' is an equilateral triangle and I,I' the midpoints of BE,BE', resp. Denote:

F := AE /\ DC, F' := AE' /\ DC

M := IF /\ DE, M' := I'F' /\ D'E'

K := IF /\ I'F'

N := AD /\ IF, N' := AD /\ I'F'

The triangles IEM, I'E'M', KII', DMN, DM'N', KNN' are equilateral.

Antreas

SQUARE PROBLEM

Let ABCD be a square. E is a variable point on the line BC, I is the midpoint of BE. CD and AE meet at F, IF and DE meet at M. The locus of M is the circumcircle of the square ABCD.

Bernard Gibert, Hyacinthos Message #19822

Solution:

To prove that M lies on the circumcircle of ABCD is enough to prove that the quadrilateral BMCD is cyclic, and since DCB = 90 d., to prove that EMB = 90 d. If triangle EMB is right angled, then since IB = IE ==> IB = IE = IM.

I will prove that IE = IM.

I take the point E between B,C (if E is on the extension of BC, then signs in the calculations are changed).

Denote AB = BC = CD = DA := a and EC := x

We have:

IE = (a-x)/2

From the similar triangles FAD,FEC we get:

CF = ax / (a-x)

DF = a^2 / (a-x)

In the right triangle ICF we have:

IF^2 = IC^2 + CF^2 = (IE + EC)^2 + CF^2 = (a^2 + x^2) / 2(a-x)

Now, by Menelaus Theorem in the triangle ICF with transversal MED, we get:

MI/MF . DF/DC . EC/EI = 1

==>

[IM /(IF-IM)]. DF/DC . EC/EI = 1

==>

(IM / [[(a^2+x^2) / 2(a-x)] - IM]) . (a^2 / (a-x)]/a) . (x /[(a-x)/2]) = 1

==> IM = (a-x)/2

So IE = IM,

QED

Antreas

TRIANGLE CONSTRUCTION A, a, h_a + h_b + h_c

To construct triangle ABC if are given A, a, h_a + h_b + h_c (sum of altitudes)

We have:

A and a known ==> R is known.

h_a + h_b + h_c = 2R(bc + ca + ab)

==> bc + ca + ab is known.

Solution 1:

bc + ca + ab = a(b+c) + bc := k^2 (known)

a^2 = b^2 + c^2 - 2bc.cosA = (b+c)^2 - 2bc(1 + cosA) = (b+c)^2 - 4bc(cos(A/2))^2

Denote b+c := X, bc := Y^2

==>

aX + Y^2 = k^2

a^2 = X^2 - 4Y^2(cos(A/2)^2

==> X^2 + 4a(cos(A/2)^2.X - 4k^2(cos(A/2)^2 - a^2 = 0

==> b+c is known and also bc is known.

==> b,c are known.

Solution 2:

We have:

2(bc + ca + ab) = (b + c)^2 - (b^2 + c^2) + 2a(b + c) (1)

Let ABC be the triangle in question. The bisector AD of A intersects the circumcircle at E. Let EF be the diameter perpendicular to BC at its midpoint M (see figure).

Denote:

AE = d, AM = m_a,

EB = EC = m, known

AF = y

EM = x, MF = 2R - x = z, known

(since the isosceles triangles EBC, FBC are known: BC = a and have known angles.)

We have:

y^2 = (2R)^2 - d^2 (from the right triangle AEF)) (2)

m(b + c) = ad (by Ptolemy Theorem in the cyclic quadril. ABEC)

==> b + c = am / d (3)

b^2 + c^2 = 2(m_a)^2 + (a^2 / 2) (Theorem of median in ABC) (4)

zd^2 + xy^2 = 2R(m_a)^2 + 2Rxz (by Stewart Theorem in AFE) (5)

(2) and (5) ==> zd^2 + x((2R)^2 - d^2) = 2R(m_a)^2 + 2Rxz (6)

(1) and (3), (4), (6) ==>

2(bc + ca + ab) = (a/m)^2d^2 - ((z-x)/R)d^2 - 4Rx + 2xz - (a^2)/2 + (2a^2/m)d

==>

[(z-x)/R - (a/m)^2]d^2 - (2a^2/m)d - 2xz + 4Rx + (a^2/2) + 2(bc + ca + ab) = 0

==> AE = d is known.

Construction:

We construct the isosceles triangle EBC with BC = a, BEC = 180 - A, BE = CE. The circle (E, d) intersects the circumcircle of EBC at A.

TRIANGLE CONSTRUCTION A, h_b+a, h_c+a

To construct triangle ABC if are given A, h_b + a, h_c + a, where h_b,h_c are the altitudes from B,C, resp.

Solution 1

We have:

2Rh_b = ac
2Rh_c = ab

==>

h_b + a = (ac/2R) + a = (a/2R).(2R + c)

h_c + a = (ab/2R) + a = (a/2R).(2R + b)

We have that angle A is known ==> a/2R is known

[Geometric Proof:

Let O be the circumcenter and M the midpoint of BC.
The triangle BOC has known angles since A is known ==>

MC / OC is known ==> (a/2)/R = a/2R is known].

So the problem is equivalent to construct triangle if are given A, 2R + b, 2R + c.

Analysis:

Let ABC be the triangle in question, AOD diameter of the circumcircle and B',C' points on the extensions of AB,AC such that BB' = CC' = AD [= 2R].

A is known

AB' = AB + BB' = c + 2R, known

AC' = AC + CC' = b + 2R, known

==> The triangle AB'C' can be constructed.

We have:

DB is perpendicular to AB' (since AD is diameter) and BB' = AD [=2R]

==> the locus of D is the parabola with focus A and directrix the perpendicular to AB' at B' (see LEMMA).

Similarly:

DC is perpendicular to AC' and CC' = AD ==> the locus of D is the parabola with focus A and directric the perpendicular to AC' at C'.

Therefore D is intersection point of the two loci. B,C are the (other than A) intersections of the circle of diameter AD with the lines AB',AC' resp.

LEMMA:

In triangle ABC, let BC be fixed and D the orthogonal projection of A on BC. If BD = AC then the locus of A is a parabola.

Let A' be the orthogonal projection of A on the perpendicular to BC at B. We have AC = BD and BD = AA' ==> AC = AA' ==> the locus of A is the parabola with focus C and directrix the perpendicular to BC at B.

Solution 2

Let BB' = h_b, CC' = h_c be the altitudes from B,C, resp.

The right triangles C'AC,B'AB are similar and have known angles (since A is known)

==>

CC' / CA = BB' / BA is known ==>

h_c / b = h_b / c = (h_c - h_b) / (b - c) = [(h_c + a) - (h_b + a)] / (b - c)

==> b - c is known.

So the problem is equivalent to construct triangle if are given A, b - c, h_c + a.

Analysis:

Let ABC be the triangle in question with AC > AB.

Let D be the point on AC between A and C such that AD = AB, CE the altitude from C, and Z the intersection of the lines BD and CE.

In the triangle CDZ we have:

CD = AC - AD = b - c, known.

Angles (BDC) = (DAB) + (DBA) = A + (90 - (A/2)) = 90 + (A/2), known

(DCZ) = 90 - (CAE) = 90 - A, known.

Therefore the triangle CDZ can be constructed.

Let H be the point on the extension of CE such that EH = BC.

We have CH = CE + EH = h_c + a, known.

BE is perpendicular to CH and BC = EH ==> (according to LEMMA) the locus of B is the parabola with focus C and directrix the perpendicular to CH at H.

So B is the intersection of the line DZ and the parabola.

Solution 3

Let ABC be the triangle in question and O its circumcenter.

The perpendicular bisector of BC intersects the circumcircle at D,E (as in the figure).

Denote:

CD = DB =: m

EC = EB =: n

AD := d

EA := e

We have:

b + 2R := k1, known (1)

c + 2R := k2, known (2)

m(b + c)= ad (3)(by Ptolemy Theorem in the cyclic quadrilateral ABDC)

The triangle BCD has known angles (DCB = DBC = A/2, CDB = 180 - A)

==> a/m := t is known.

==> b + c = (a/m)d = td

ae + cn = bn (4)(by Ptolemy Theorem in the cyclic quadrilateral ABCE) ==>
e = (b - c)n/a

We have:

b - c = k1 - k2, known

n/a is known since the triangle CEB has known angles (CEB = A, ECB = EBC = 90 - (A/2))

Therefore e = (b - c)n/a is known.

EA^2 = AD^2 - AD^2 (5)(by Pythagorean Theorem in the right triangle ADE)or e^2 = 4R^2 - d^2, known.

Now, from:

(1) and (2) ==> b + c = k1 + k2 - 4R (6)

(6) and (3) ==> d = (k1 + k2 - 4R) / t (7)

(7) and (5) ==> 4R^2 - ((k1 + k2 - 4R)/t)^2 = e^2

== > R is known.

So the problem is equivalent to construct triangle if are given A, b - c, R or A, b - c, a (the solution is left to the reader).

Exercises:

To construct triangle ABC if are given:

1. A, h_b - a, h_c + a

2. A, h_b - a, h_c - a

TRIANGLE CONSTRUCTION A, 2b+a, 2c+a

To construct triangle ABC if are given A, 2b + a = m, 2c + a = n

Analysis

Let ABC be the triangle in question.

We have:

m+n = 2(a+b+c) = 4s ==> the semiperimeter s is known

m-n = 2(b-c) ==> the difference b-c is known

Let E,D the points the a-excircle (Ia) touches AC,BC, resp.
The triangle DAIa has:
ADIa = 90 d., DAIa = A/2, AD = s. Therefore IaD = IaE = r_b is known.

Let M be the midpoint of BC. We have BIaC = 90-(A/2) and ME = (|b-c|)/2.(So the problem is eqivalent to construct triangle if are given:
A, b-c, r_b)

IaM^2 = IaE^2 + ME^2 = (r_b)^2 + ((b-c)/2)^2 ==> the median IaM is known.

In the triangle IaBC we know the angle Ia, the altitude and the median from Ia, therefore the problem is equivalent to construct triangle if are given:

A, h_a, m_a (altitude, median from A, resp.). This construction is left to the reader.

Exercises:

To construct triangle ABC if are given:

1. A, 2b - a = m, 2c + a = n

2. A, 2b - a = m, 2c - a = n

TRIANGLE CONSTRUCTION A, a + b, a + c

To construct triangle ABC if are given A, a + b, a + c

Solution 1.

Analysis:

Let ABC be the triangle in question. Let D, E be two points on the extensions of AC,AB, resp. such that CD = BE = BC = a.

The parallel from A to BC intersects DB at Z.

The triangle AZB is similar to triangle CDB ==>

AZ / AD = CB / CD = 1 ==> AZ = AD (1)

The parallel from Z to AE intersects DE at Q.

The triangles AZQ and CBE are similar ==>

AZ / ZQ = CB / BE = 1 ==> ZQ = AZ (2)

(1) /\ (2) ==> AD = AZ = ZQ (3).

The parallel from Z to EQ intersects AB at H.

We have HE = ZQ (4) (since EHZQ is parallelogram)

(3) /\ (4) ==> HE = AD.

Construction:

I construct the triangle ADE such that AD = b+a, AE = c+a, angle DAE = A.

Let H be on AE such that EH = AD.

The circle (A, AD) intersects the parallel from H to DE at Z.

The line DZ intersects AE at B. The parallel from B to AZ intersects AD at C.

ABC is the required triangle.

The Proof and Investigation are left to the reader.

Reference:
Ioannis Panakis: Solutions of the Exercises of the MATHEMATICS of the 5th Class of Greek Gymnasium, vol I, Athens, Kokotsakis Bookstore, p. 90.

Solution 2.

Analysis:

Let ABC be the triangle in question. Let B', C' be two points on the extensions of AB,AC, resp. such that BB' = CC' = BC = a.

Let D be the intersection of the lines: Parallel from B' to BC and Parallel from C to BB'. The quadrilateral BCDB' is rhombus. The isosceles triangle DCC' has fixed angles: DCC' = A, CDC' = CC'D = (90-A)/2 ("remains similar to itself"). Therefore CD/C'D is fixed, and since CD = B'D ==> B'D/C'D is fixed.

So the point D lies on a known line forming with AC' angle (90-A)/2 and on the Apollonius circle (B'C', B'D/C'D).

The Construction, Proof and Investigation are left to the reader.

Reference:
EUCLID [publ. by the Greek Mathematical Society], December 1982.

Solution 3.

Analysis:

Let ABC be the triangle in question. Let B', C' be two points on the extensions of AB,AC, resp. such that BB' = CC' = BC = a.

Let D be the intersection of BC' and CB' and E the intersection of the parallel from C' to AB' and the parallel from B' to BC'.

The isosceles trianle CC'E has fixed angles:

(CC'E) = 180 - A, (C'CE) = (C'EC) = A/2.

We have:

Angle (CB'E) = (CDC') = (CBD) + (BCD) = C/2 + B/2 = 90 - (A/2) : fixed.

Construction:

I construct the triangle AB'C' such that AB' = c+a, AC' = b+a, angle (B'AC') = A. Let L be an arbitrary point on C'A and M a point on the parallel through C' to AB' such that C'L = C'M. The circle (LM, 90-(A/2)), ie the circle with chord LM and angle 90-(A/2), intersects B'C at N. The parallel through B' to LN intersects AC' at C. The parallel through B' to MN intersects C'M at E. The parallel through C' to B'E intersects AB' at B. The triangle ABC is the required triangle.

Proof (to prove B'B = BC = CC') / Investigation: Left to the reader.

Reference: A. P. Hatzipolakis (1982)

Exercise:
Contruct the triangle if are given:
1. A, b + a, c - a
2. A, b - a, c - a

TRIANGLE RESOLUTION A, a+b-c, R+r

Problem:
To resolve triangle ABC if are given A, a+b-c, R+r
See Hyacinthos Message #19749

Resolution:

We have:

sinA + sinB - sinC = 4sin(A/2)sin(B/2)cos(C/2)

r = 4Rsin(A/2)sin(B/2)sin(C/2)

==>

a+b-c = 2R(sinA+sinB-sinC) = 8Rsin(A/2)sin(B/2)cos(C/2)

R+r = R(1+4sin(A/2)sin(B/2)sin(C/2))

==>

t := (R+r)/(a+b-c) =

= (1+4sin(A/2)sin(B/2)sin(C/2))/8sin(A/2)sin(B/2)cos(C/2)

==>

4sin(A/2)sin(B/2)[sin(C/2) - 2tcos(C/2)] + 1 = 0

and since sin(B/2) = sin(90 - ((C+A)/2)) = cos((C+A)/2) =

= cos(C/2)cos(A/2) - sin(C/2)sin(A/2), ==>

4sin(A/2)[cos(A/2) + 2tsin(A/2)]sin(C/2)cos(C/2) - 4(sin(A/2)^2(sin(C/2))^2 - 8tsin(A/2)cos(A/2)(cos(A/2)^2 + 1 = 0.

We have the system of equations:

fx^2 + gy^2 + hxy + 1 = 0

x^2 + y^2 = 1

where x, y stand for the unknown sin(C/2),cos(C/2) and f,g,h are known coefficients. From these equations we get the equation:

Lx^4 + Mx^2 + N = 0, where L,M,N are known coefficients.

This equation has constructible roots [Read THIS], therefore the problem has a constructible Euclidean solution (ie by ruler and compass)