Eight Solved and Eight Open Problems in Elementary Geometry

Eight Solved and Eight Open Problems in Elementary Geometry Florentin Smarandache Math & Science Department University of New Mexico, Gallup, USA In this paper we review eight previous proposed and solved problems of elementary 2D geometry [4], and we extend them either from triangles to polygons or from 2D-space to 3Dspace and make some comments about them.

Problem 1.

We draw the projections M i of a point M on the sides Ai Ai +1 of the polygon A1... An .

Prove that: 2 2 2 2 2 M 1 A1 + ... + M n An = M 1 A2 + ... + M n −1 An + M n A1 Solution 1.

For all i we have: 2 2 2 2 MM i = MAi − Ai M i = MAi +1 − Ai +1M i It results that: 2 2 2 2 M i Ai − M i Ai +1 = MAi − MAi +1 From where:

∑( M A i

i

i

2

− M i Ai +1

2

) = ∑ ( MA

i

i

2

− MAi +1

2

2

)=0

Open Problem 1.

1.1. If we consider in a 3D-space the projections M i of a point M on the edges Ai Ai +1 of a polyhedron A1... An then what kind of relationship (similarly to the above) can we find?

1.2. But if we consider in a 3D-space the projections M i of a point M on the faces Fi of a polyhedron A1... An with k≥4 faces, then what kind of relationship (similarly to the above) can we find?

Problem 2.

Let’s consider a polygon (which has at least 4 sides) circumscribed to a circle, and D the set of its diagonals and the lines joining the points of contact of two non-adjacent sides. Then D contains at least 3 concurrent lines.

Solution 2.

Let n be the number of sides. If n = 4 , then the two diagonals and the two lines joining the points of contact of two adjacent sides are concurrent (according to Newton's Theorem).

The case n > 4 is reduced to the previous case: we consider any polygon A1... An (see the figure)

P

Ai+1

Ai

B1

Aj-1

B2

O

Ai-1 R

B4 Aj

B3

Aj+1

circumscribed to the circle and we choose two vertices Ai , A j ( i ≠ j ) such that A j A j −1 ∩ Ai Ai +1 = P

and

A j A j +1 ∩ Ai Ai −1 = R

.

Let Bh , h ∈ {1, 2, 3, 4} the contact points of the quadrilateral PA j RAi with the circle of center O . Because of the Newton’s theorem, the lines Ai A j , B1 B3 and B2 B4 are concurrent.

Open Problem 2.

2.1. In what conditions there are more than three concurrent lines?

2.2. What is the maximum number of concurrent lines that can exist (and in what conditions)?

2.3. What about an alternative of this problem: to consider instead of a circle an ellipse, and then a polygon ellipsoscribed (let’s invent this word, ellipso-scribed, meaning a polygon whose all sides are tangent to an ellipse which inside of it): how many concurrent lines we can find among its diagonals and the lines connecting the point of contact of two non-adjacent sides?

2.4. What about generalizing this problem in a 3D-space: a sphere and a polyhedron circumscribed to it?

2.5. Or instead of a sphere to consider an ellipsoid and a polyhedron ellipsoido-scribed to it?

Of course, we can go by construction reversely: take a point inside a circle (similarly for an ellipse, a sphere, or ellipsoid), then draw secants passing through this point that intersect the

circle (ellipse, sphere, ellipsoid) into two points, and then draw tangents to the circle (or ellipse), or tangent planes to the sphere or ellipsoid) and try to construct a polygon (or polyhedron) from the intersections of the tangent lines (or of tangent planes) if possible.

For example, a regular polygon (or polyhedron) has a higher chance to have more concurrent such lines.

In the 3D space, we may consider, as alternative to this problem, the intersection of planes (instead of lines).

Problem 3.

In a triangle ABC let’s consider the Cevians AA ', BB ' and CC ' that intersect in P . Calculate the minimum value of the expressions: PA PB PC E ( P) = + + PA ' PB ' PC ' and PA PB PC F ( P) = ⋅ ⋅ PA ' PB ' PC ' where A ' ∈ [ BC ] , B ' ∈ [CA] , C ' ∈ [ AB ]

.

Solution 3.

We’ll apply the theorem of Van Aubel three times for the triangle ABC , and it results: PA AC ' AB ' = + PA ' C 'B B 'C PB BA ' BC ' = + PB ' A 'C C'A PC CA ' CB ' = + PC ' A' B B'A

If we add these three relations and we use the notation AC ' C 'B

= x>0,

AB ' B 'C

= y > 0,

BA ' A 'C

=z>0

then we obtain: ⎛ 1⎞ ⎛ 1⎞ ⎛ 1⎞ E ( P) = ⎜ x + ⎟ + ⎜ x + ⎟ + ⎜ z + ⎟ ≥ 2 + 2 + 2 = 6 y⎠ ⎝ y⎠ ⎝ z⎠ ⎝ The minimum value will be obtained when x = y = z = 1 , therefore when P will be the gravitation center of the triangle.

When we multiply the three relations we obtain

⎛ 1⎞ ⎛ 1⎞ ⎛ 1⎞ F ( P) = ⎜ x + ⎟ ⋅ ⎜ x + ⎟ ⋅ ⎜ z + ⎟ ≥ 8 y⎠ ⎝ y⎠ ⎝ z⎠ ⎝ Open Problem 3.

3.1.

Instead of a triangle we may consider a polygon A1A2…An and the lines A1A1’, A2A2’, …, AnAn’ that intersect in a point P. Calculate the minimum value of the expressions: E ( P) =

F ( P) =

3.2.

PA1 PA2 PAn + + ... + PA1 ' PA2 ' PAn ' PA1 PA2 PAn ⋅ ⋅ ... ⋅ PA1 ' PA2 ' PAn '

Then let’s generalize the problem in the 3D space, and consider the polyhedron A1A2…An and the lines A1A1’, A2A2’, …, AnAn’ that intersect in a point P. Similarly, calculate the minimum of the expressions E(P) and F(P).

Problem 4.

If the points A1 , B1 , C1 divide the sides BC, CA respectively AB of a triangle in the same ratio k > 0, determine the minimum of the following expression: 2

2

AA1 + BB1 + CC1

2

.

Solution 4.

Suppose k > 0 because we work with distances.

BA1 = k BC , CB1 = k CA , AC1 = k AB We’ll apply tree times Stewart’s theorem in the triangle ABC , with the segments AA1 , BB1 , respectively CC1 : AB ⋅ BC (1 − k ) + AC ⋅ BC k − AA1 ⋅ BC = BC 2

2

2

3

(1 − k ) k

where AA1

2

= (1 − k ) AB + k AC − (1 − k ) k BC

2

BB1

2

= (1 − k ) BC + k BA − (1 − k ) k AC

2

CC1

2

= (1 − k ) CA + k CB − (1 − k ) k AB

2

2

2

similarly, 2

2

By adding these three equalities we obtain: 2

2

AA1 + BB1 + CC1

2

2

2

(

= ( k 2 − k + 1) AB + BC + CA 2

2

2

),

1 , which is the case when the three lines from the 2 enouncement are the medians of the triangle.

3 2 2 2 The minimum is AB + BC + CA .

4

which takes the minimum value when k =

(

)

Open Problem 4.

4.1. If the points A1’, A2’, …, An’ divide the sides A1A2, A2A3, …, AnA1 of a polygon in the same ratio k>0, determine the minimum of the expression: 2 2 2 A1 A1 ' + A2 A2 ' + ... + An An ' .

4.2. Similarly question if the points A1’, A2’, …, An’ divide the sides A1A2, A2A3, …, AnA1 in the positive ratios k1, k2, …, kn respectively.

4.3. Generalize this problem for polyhedrons.

Problem 5.

In the triangle ABC we draw the lines AA1 , BB1 , CC1 such that 2

2

2

2

2

2

A1 B + B1C + C1 A = AB1 + BC1 + CA1 .

In what conditions these three Cevians are concurrent?

Partial Solution 5.

They are concurrent for example when A1, B1, C1 are the legs of the medians of the triangle BCA. Or, as Prof. Ion Pătrașcu remarked, when they are the legs of the heights in an acute angle triangle BCA .

More general.

The relation from the problem can be written also as: a ( A1 B − A1C ) + b ( B1C − C1 A ) + c ( C1 A − C1 B ) = 0 , where a, b, c are the sides of the triangle.

We’ll denote the three above terms as α , β , and respective γ , such that α + β + γ = 0 . α α = a ( A1 B − A1C ) ⇔ = A1 B − A1C − 2 A1C a where a2 a a 2a 2 2a 2 − a 2 + α a − A1C α a − 2 A1C = ⇔ 2 = ⇔ = 2 ⇔ = a2 a a − α 2 A1C 2 A1C a −α a2 − α A1C Then

a2 + α . = 2 A1C a −α A1 B

Similarly: C1 A c 2 + γ b2 + β and = B1 A b 2 − β C1 B c 2 − γ In conformity with Ceva’s theorem, the three lines from the problem are concurrent if and only if: A1 B B1C C1 A ⋅ ⋅ = 1 ⇔ ( a 2 + α )( b 2 + β )( c 2 + γ ) = ( a 2 − α )( b 2 − β )( c 2 − γ ) A1C B1 A C1 B B1C

=

Unsolved Problem 5.

Generalize this problem for a polygon.

Problem 6.

In a triangle we draw the Cevians AA1 , BB1 , CC1 that intersect in P . Prove that PA PB PC AB ⋅ BC ⋅ CA ⋅ ⋅ = PA1 PB1 PC1 A1 B ⋅ B1C ⋅ C1 A Solution 6.

In the triangle ABC we apply the Ceva’s theorem: AC1 ⋅ BA1 ⋅ CB1 = − AB1 ⋅ CA1 ⋅ BC1 (1) In the triangle AA1 B , cut by the transversal CC1 , we’ll apply the Menelaus’ theorem: AC1 ⋅ BC ⋅ A1 P = AP ⋅ A1C ⋅ BC1 (2) In the triangle BB1C , cut by the transversal AA1 , we apply again the Menelaus’ theorem: A C1

P

B1 C

B

A1

BA1 ⋅ CA ⋅ B1 P = BP ⋅ B1 A ⋅ CA1 (3) We apply one more time the Menelaus’ theorem in the triangle CC1 A cut by the transversal BB1 : AB ⋅ C1 P ⋅ CB1 = AB1 ⋅ CP ⋅ C1 B (4) We divide each relation (2), (3), and (4) by relation (1), and we obtain:

PA BC B1 A = ⋅ PA1 BA1 B1C PB CA C1 B = ⋅ PB1 CB1 C1 A PC AB A1C = ⋅ PC1 AC1 A1 B Multiplying (5) by (6) and by (7), we have: PA PB PC AB ⋅ BC ⋅ CA AB1 ⋅ BC1 ⋅ CA1 ⋅ ⋅ = ⋅ PA1 PB1 PC1 A1 B ⋅ B1C ⋅ C1 A A1 B ⋅ B1C ⋅ C1 A but the last fraction is equal to 1 in conformity to Ceva’s theorem.

(5) (6) (7)

Unsolved Problem 6.

Generalize this problem for a polygon.

Problem 7.

Given a triangle ABC whose angles are all acute (acute triangle), we consider A ' B ' C ' , the triangle formed by the legs of its altitudes.

In which conditions the expression: A ' B ' ⋅ B 'C ' + B 'C ' ⋅ C ' A ' + C ' A ' ⋅ A ' B ' is maximum?

A C’

b-y B’

z

y c-z B

x

C

’

A

a-x

Solution 7.

We have

ΔABC ~ ΔA ' B ' C ' ~ ΔAB ' C ~ ΔA ' BC '

We note BA ' = x, CB ' = y , AC ' = z .

It results that

(1)

∧

A ' C = a − x, B ' A = b − y , C ' B = c − z

∧

∧

∧

∧

∧

∧

∧

∧

BAC = B ' A ' C = BA ' C '; ABC = AB ' C ' = A ' B ' C '; BCA = BC ' A ' = B ' C ' A From these equalities it results the relation (1) A'C ' x (2) = ΔA ' BC ' ~ ΔA ' B ' C ⇒ a−x A' B ' ΔA ' B ' C ~ ΔAB ' C ' ⇒

A'C ' c−z = z B 'C '

(3)

B 'C ' b− y (4) = y A' B ' From (2), (3) and (4) we observe that the sum of the products from the problem is equal ΔAB ' C ' ~ ΔA ' B ' C ⇒

to: 2

2

2

1 a⎞ ⎛ b⎞ ⎛ c⎞ ⎛ x ( a − x ) + y (b − y ) + z ( c − z ) = ( a 2 + b2 + c2 ) − ⎜ x − ⎟ − ⎜ y − ⎟ − ⎜ z − ⎟ 4 2⎠ ⎝ 2⎠ ⎝ 2⎠ ⎝ a b c which will reach its maximum as long as x = , y = , z = , that is when the altitudes’ legs are 2 2 2 in the middle of the sides, therefore when the ΔABC is equilateral. The maximum of the 1 expression is ( a 2 + b 2 + c 2 ) .

4 1 Conclusion : If we note the lengths of the sides of the triangle Δ ABC by ||AB|| = c, ||BC|| = a, ||CA|| = b, and the lengths of the sides of its orthic triangle Δ A`B`C` by ||A`B`|| = c`, ||B`C`|| = a`, ||C`A`|| = b`, then we proved that:

4(a`b` + b`c` + c`a`) ≤ a2 + b2 + c2.

Unsolved Problems 7.

7.1. Generalize this problem to polygons. Let A1A2…Am be a polygon and P a point inside it.

From P, which is called a pedal point, we draw perpendiculars on each side AiAi+1 of the polygon and we note by Ai’ the intersection between the perpendicular and the side AiAi+1. Let’s extend the definition of pedal triangle to a pedal polygon in a straight way: i.e. the polygon formed by the orthogonal projections of a pedal point on the sides of the polygon. The pedal polygon A1’A2’…Am’ is formed. What properties does this pedal polygon have?

7.2. Generalize this problem to polyhedrons. Let A1A2…An be a polyhedron and P a point inside it. From P we draw perpendiculars on each edge AiAj of the polyhedron and we note by Aij’ the intersection between the perpendicular and the side AiAij. Let’s name the

1

This is called the Smarandache’s Orthic Theorem (see [2], [3]).

new formed polyhedron an edge pedal polyhedron, A1’A2’…An’. What properties does this edge pedal polyhedron have?

7.3. Generalize this problem to polyhedrons in a different way. Let A1A2…An be a poliyhedron and P a point inside it. From P we draw perpendiculars on each polyhedron face Fi and we note by Ai’ the intersection between the perpendicular and the side Fi. Let’s call the new formed polyhedron a face pedal polyhedron, which is A1’A2’…Ap’, where p is the number of polyhedron’s faces. What properties does this face pedal polyhedron have?

Problem 8.

Given the distinct points A1 ,..., An on the circumference of a circle with the center in O and of ray R .

180o Prove that there exist two points Ai , A j such that OAi + OAj ≥ 2 R cos n Solution 8.

Because A1OA2 + A2OA3 + ... + An −1OAn + An OA1 = 360o and ∀i ∈ {1, 2,..., n} ,

Ai OAi + 2 > 0o , it result that it exist at least one angle

(otherwise it follows that S >

360o n

3600 ⋅ n = 360o ). n

An-1 An

AOA i j ≤

Aj O

A1

M Ai

A2 OAi + OAj = OM ⇒ OAi + OA j = OM 360o The quadrilateral OAi MA j is a rhombus. When α is smaller, OM is greater. As α ≤ , it n α 180o . results that: OM = 2 R cos ≥ 2 R cos 2 n

Open Problem 8: Is it possible to find a similar relationship in an ellipse? (Of course, instead of the circle’s radius R one should consider the ellipse’s axes a and b.)

References: [1] Cătălin Barbu, Teorema lui Smarandache, in his book “Teoreme fundamentale din geometria triunghiului”, Chapter II, Section II.57, p. 337, Editura Unique, Bacău, 2008.

[2] József Sándor, Geometric Theorems, Diophantine Equations, and Arithmetic Functions, AR Press, pp. 9-10, Rehoboth 2002.

[3] F. Smarandache, Eight Solved and Eight Open Problems in Elementary Geometry, in arXiv.org at http://arxiv.org/abs/1003.2153 .

[4] F. Smarandache, Problèmes avec et sans… problèmes!, pp. 49 & 54-60, Somipress, Fés, Morocoo, 1983.