INRADIUS 5

Let ABC be a triangle and D a point on AB such that inradius of CDA = inradious of CDB := d.

Find the primitive heronian right triangles ABC, A = 90 d. (ie a^2 = b^2 + c^2, a,b,c integers, gcd(a,b,c) = 1) with d integer.

ABC primitive heronian right triangle with A = 90 d. ==>

a = x^2 + y^2

b or c = x^2 - y^2

c or b = 2xy

where x > y > 0 integers.

We have inradius r = y(x-y) and c = 2d^2 / (2d - r) [see HERE].

1. c = 2xy, b = x^2 - y^2

==> xy = d^2 /(2d - y(x-y)) ==> d = xy +-ysqrt(xy) and since x>y,

d = xy + ysqrt(xy).

d integer ==> xy = z^2 and since gdc(x,y) = 1 ==> x = m^2, y = n^2, where m,n are integers.

Therefore
a = m^4 + n^4, b = m^4 - n^4, c = 2m^2*n^2, d = mn^2*(m+n)

2. c = x^2 - y^2, b = 2xy

==>

x^2 - y^2 = 2d^2 / (2d - y(x-y)) ==>

2d = x^2 - y^2 +- ((x-y)sqrt(x^2-y^2)) and since d is positive,

2d = x^2 - y^2 + ((x-y)sqrt(x^2-y^2))

d integer ==> x^2 - y^2 = z^2 ==> x = m^2 + n^2, y = m^2 - n^2 or 2mn

2.1 : x = m^2 + n^2, y = m^2 - n^2

==>

a = (m^2+n^2)^2 + (m^2-n^2)^2

b = 2(m^2+n^2)(m^2-n^2)

c = (m^2+n^2)^2 - (m^2-n^2)^2

d = 2mn^2*(m+n)

2.2 : x = m^2 + n^2, y = 2mn

==>

a = (m^2+n^2)^2 + (m^2-n^2)^2

b = 4mn(m^2+n^2)

c = (m^2-n^2)^2

d = m(m+n)(m-n)^2

GOLDEN ISOSCELES TRIANGLE

Let ABC be an isosceles triangle with AB = AC and A'B'C'D' the inscribed square in ABC based on AC (A' on AB, B' on BC, C' on CA near C and D' on CA near A). Let y be the side of the square inscribed in BA'C' based on A'B' and z the side of the square inscribed in the right triangle D'A'A based on AD'.

If y = z, find the angles of the triangle.

Solution:

Let x be the side of the square A'B'C'D' and h the altitude BB*.

we have:

x = bh / (b+h)

y = bh^2 / (b+h)^2 = x^2 / b

z = xAD' / (x + AD') = x^2cotA /(x + xcotA) = xcotA / (1 + cotA)

y = z ==> x / b = cotA / (1 + cotA) ==>

(h/b) / (1 + (h/b)) = cotA / (1 + cotA) ==>

(h/c) / (1 + (h/c)) = cotA / (1 + cotA) ==>

sinA / (1 + sinA) = cotA / (1 + cotA) ==>

sinA = cotA ==> (cosA)^2 + cosA - 1 = 0 ==>

cosA = (-1 + sqrt(5))/2 (since 0 < A < 180) ==> A = Αrccos(1/φ) and B = C = 90 - (A/2)

Note: B* divides AC in golden ratio.

Addendum (13 April):

There is one more solution for A > 90 d.

See Hyacinthos Message 19985

REGULAR POLYGON PROBEM

Let A1A2A3... An be a regular n-gon. The perpendicular
to A1A2 at A2 intersects A3A4 at K. The parallel through
A3 to KA1 intersects A1A2 at M.

For which n's the point M is the midpoint of A1A2?

APH, Hyacinthos message 19937

Trigonometric Solution:

Let: A1A2 = A2A3 =... = 1, angles(A1A2A3) = (A2A3A4) = ... = ω

Denote: angle(KA1A2) = (A3MA2) = θ, angle(MA3A2) = φ, KA2 := d

We have:

In the right triangle A2KA1: tanθ = d

In the triangle KA2A3: d/sinω = 1/sin(270-2ω) ==> d / sinω = - 1/cos2ω

In the triangle MA2A3: (1/2)/sinφ = 1/sinθ ==> 2sinφ = sinθ
and since φ + θ = 180 - ω, ==> 2sin(ω+θ) = sinθ

So we have the system of two equations:

tanθ = - sinω/cos2ω

2sin(ω+θ) = sinθ

==>

2cos2ω - 2cosω + 1 = 0 ==> 4(cosω)^2 - 2cosω - 1 = 0 ==>

cosω = (1 +- sqrt5) / 4 = cos36 or cos108

We have:
180 > ω = 180(n-2)/n >= 60 since n >= 3 ==> ω = 108 d. and n = 5.

Other solutions by Nikos Dergiades, Hyacinthos messages
19939,19946

INRADIUS 4 (Golden Ratio)

Let ABC be a triangle and Aa a point on BC between B,C.

Denote:

h_a : the A-altitude of ABC

r_a : the A-exradius of ABC

r_ab, r_ac : the A-exradii of AAaB and AAaC.

We have:

h_a = 2r_ab.r_ac / (r_a - (r_ab + r_ac)) (1)

[See HERE]

Now, assume that r_ab = r_ac and r_a = 2h_a.

Denote r_ab / h_a := x

(1) ==>

1 = 2x^2 / (2 - 2x) ==> x^2 + x - 1 = 0

==>

0 < x = (- 1 + sqrt5)/2 = 1/φ, where φ is the "golden number" φ = (1 + sqrt5)/2.

Golden Section:

Let Iab, Iac be the A-excenters of AAbB, AAcC, resp., and A* the intersection of the line IabIac and the line of AA' (of the A-altitude).

A' divides ΑA* in golden ratio.

Addendum:

If the triangle ABC is isosceles AB = AC, then the foot A' of the A-altitude coincides with the point of contact of the A-excircle and BC:

CONSTRUCTION:

Let (A),(B) be two externally tangent circles at C with radius of (A) = 4.radius of (B). The line ACB intersects again the circle (B) at D. A tangent from D to circle (A) intersects the common internal tangent of (A),(B) at E. The bisector of the right angle ECA intersects EA at F. Let G be the orthogonal projection of F on AB. C divides GD in Golden Ratio.

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.