## Πέμπτη, 10 Φεβρουαρίου 2011

### 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'

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

Antreas

## Δευτέρα, 7 Φεβρουαρίου 2011

### 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

## Τρίτη, 1 Φεβρουαρίου 2011

### 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.