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Solution 3

([Honsberger, p. 26] ascribes this solution to Torricelli himself and mentions that it was rediscovered almost 300 years later by F. Riesz. In a private communication, Douglas Rogers mentioned that the solution has been also proposed by J. Steiner. Thus the same object that is most frequently referred to as Fermat's point, Torricelli's point and Fermat-Torricelli point, is sometimes also named after J. Steiner, as Steiner's point, especially in the framework of Steiner's networks. [Tikhomirov, p. 31] and [Courant and Robbins, p. 354] even refer to Fermat's problem as Steiner's. [Johnson, p. 221] attributes the present solution to Steiner.)

The following solution depends on explicit knowledge of the fact that, for Fermat's point, AFB = BFC = AFC = 120o. Without proving that such a point exists, it shows that, if it does, it solves Fermat's problem.

At the vertices of ΔABC draw perpendiculars to lines AF, BF, and CF. The new lines form an equilateral triangle RST. Let F1 be a point different from F. Drop perpendiculars F1A1, F1B1 and F1C1 to the sides of ΔRST. It's a nice property (known as Viviani's theorem) of equilateral triangles that in ΔRST,
 FA + FB + FC = F1A1 + F1B1 + F1C1.

(The sum of the distances from a point inside an equilateral triangle to the sides of the triangle does not depend on the point. )

On the other hand, obviously,

 F1A1 + F1B1 + F1C1 < F1A + F1B + F1C

All that remains is to combine the two.

(Douglas Rogers made a delightful observation. ΔRST is the largest equilateral triangle circumscribing ΔABC. According to Viviani's theorem, FA, FB, FC add up to the altitude of ΔRST. For any other equlateral ΔLMN curcumscribing ΔABC, at least one of FA, FB, or FC is not perpendicular to the side and hence is longer than such perpenduclar. It follows that the sum FA + FB + FC is bound to be larger than the alititude of ΔLMN. This implies that the side of ΔLMN is smaller than that of ΔRST. In passing, ΔRST is known as the antipedal triangle of F with respect two ΔABC.)

Solution 4

(The proof by Lou Talman is found on one of the discussions at the mathforum. Still very simple, it uses a little bit of analytic geometry and calculus. This is probably very close to one of Torricelli's original proofs.)

One can see that the Fermat point does minimize FA + FB + FC as follows: Let z be the sum FA + FB, where F is chosen to minimize FA + FB + FC, and consider the locus of points P that satisfy PA + PB = z. This locus is an ellipse with foci at A and B. Because F minimizes z + FC, the line determined by C and F must be normal to the tangent to this ellipse at F. By the reflection property of the ellipse, the angles AFC and BFC must be equal. A similar argument shows that the angles AFC (say) and AFB must also be equal. From this it follows that the three angles are all 120 degrees.

Solution 5

(The argument is similar to that in Solution 4, but from a different perspective. [F. G.-M., p. 442] credits Lhuilier (1811) with the proof.)

Assume FC is constant. Then point F that minimizes FA + FB + FC, minimizes also FA + FB and lies on the circle with center C and radius FC. The minimum is achieved for a point F on the circle for which the angles AFC and BFC are equal.

For more detail, draw a tangent to the circle at F and choose any point G on the tangent other than F. Let H be the intersection of HC with the circle. Then, assuming angles AFC and BFC equal,
 FA + FB < GA + GB < HA + HB

(The latter inequality warrants extra attention. It follows from the fact the various ellipses with foci at A and B do not intersect.) From here,
 FA + FB + FC < HA + HB + FC = HA + HB + HC.

The argument can be repeated assuming either FA or FB constant. As a result, at Fermat's point F all three angles AFC, BFC and AFB are found to be equal.

Solution 6

Fermat's point can be located with the help of Euler's generalization of Ptolemy's Theorem, see [Pedoe, pp. 93-94].

Any triangle has at least two acute angles. Given ΔABC, let angles at B and C be acute. Form an equlateral triangle BCD, with D and A on opposite sides of BC. Let (O) be the circumcircle of ΔBCD.

For a point Q, by Ptolemy's inequality

 BQ·CD + CQ·BD ≥ DQ·BC,

with equality only if Q lies on (O) and such that the quadrilateral BQCD is convex. Note that, by construction, BC = CD = BD which reduces the above to
 BQ + CQ ≥ DQ.

Therefore
 AQ + BQ + CQ ≥ AQ + DQ ≥ AD.

Unless, Q lies on AD, AQ + BQ + CQ > AD. Let P be the intersection of AD with (O) other than D. BPCD is an inscribed convex quadrilateral and P lies on AD. So in this (and only in this) case AP + BP + CP = AD. For any other selection of Q,
 AQ + BQ + CQ > AP + BP + CP.

Angle BPC is supplementary to BDC = 60° so that BPC = 120°. Further, D is in the middle of the arc BDC so that angles BPD and CPD are 60° making angles APC and APB both equal 120°.

If the triangle is acute the same construction applies to the other two sides which brings up the framework of Napoleon's theorem. The three circles intersect at Fermat's point and three lines joining the vertices of ΔABC with the opposite vertices of the Napoleon triangles concur at the point P.

The latter fact can be used for a more direct proof.

Solution 7

(This was a part of a solution by Grimbal to a problem posted at the wu::forum.)

Construct Napoleon's equilateral triangles ABC', AB'C', A'BC externally on the sides of ΔABC. Let P be the intersection of AA' and BB'. AA' is CC' rotated clockwise 60° about B, and BB' is CC' rotated counterclockwise 60° about A, it follows that angle APB' is 60°. Let X be the point on BB' making triangle APX equilateral. Now, if BB' is rotated clockwise 60° about A, then X goes to P, B' to C, and B to C'. Hence P, C, and C' are collinear; and so AA', BB', and CC' are concurrent, and any two of these lines make an angle of 120°.

Searching for the Fermat point we discovered a nice property of Napoleon's triangles. I found quite surprising generalizations of those properties in a column by David Gale in The Mathematical Intelligencer which was kindly pointed out to me by Professor McWorter.

Two lines passing through a vertex of a triangle are called isogonal with respect to that vertex if they form equal angles with its (internal) angle bisector. Following is the first generalization.

Theorem 1

As in the diagram, assume lines AB' and AC' are isogonal as are pairs CB', CA' and BA', BC'. Then three lines AA', BB', and CC' are concurrent, i.e., meet at a common point.

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Napoleon's theorem is obtained when all three angles involved are equal to 30o. The common point is then known as the First Napoleon point. (The Second Napoleon Point is obtained when the equlateral triangles are formed internally to the given triangle. Similarly, there are two Fermat's points, which are also known as the first and second isogonic centers.) If the base angles are just equal between themselves, the theorem bears the name of Ludwig Kiepert who replaced equilateral triangles with isosceles ones. An interesting specification of the theorem is obtained when the three angles add up to 180o. But the theorem also admits further generalization.

In Theorem 1, pairs of (isogonal) lines were related to angle bisectors. As is well known from elementary geometry, angle bisectors meet at a single point (incidentally, the incenter of the triangle.)

Two lines AB' and AC' through a vertex A of a triangle are said to be isotomic if they intersect the opposite side BC in points equidistant from its midpoint Ma. Theorem 1 remains valid for isotomic lines as well. And, in a certain sense, for any three concurrent lines.

A second generalization that I am about to formulate and then prove belongs to the realm of Projective Geometry.

Theorem 2

Let p, q, r be concurrent lines through the vertices A, B, and C, respectively, of ΔABC. Let PA be the pencil of lines at A and let TA be the (unique) projective mapping on PA which

1. interchanges lines AB and AC,

2. leaves p fixed.

Define PA, PB, and TA, TB similarly. For any line a in PA, let a' = TA(a). Similarly, for b in PB and c in PC, let b' = TB(b) and c' = TC(c). Let C' = a'b, B' = c'a, and A' = b'c.

Then AA', BB', and CC' are concurrent.

Note that (an apparently more general) Theorem 2 follows from Theorem 1 with a projective mapping that leaves vertices A,B,C fixed but takes the incenter into any point P (the intersection of lines p, q, r.)

To prove Theorem 2, perform the projective mapping that carries C to the origin, A to the point at infinity at y-axis, B to the point at infinity at x-axis, and P to the point (1, 1).

Lines p, q, and r are carried onto the lines x = 1, y = 1, and x = y, respectively. We thus have

 PA is the set of all vertical lines, PB is the set of all horizontal lines, PC is the set of all lines through the origin.

Transformation TA preserves the vertical direction and, therefore, is in reality a 1-dimensional projective transformation. So it's a Möbius transformation which, in general, has the form f(x) = (ax + b)/(cx + d). Transformations that interchange x = 0 with the point at infinity are given by f(x) = a + b/x (c = 1 and d = 0.) Among those, there is a single one that leaves x = 1 fixed: f(x) = 1/x. So TA maps x = a onto x = 1/a.

To simplify the notations, we will denote the vertical line with x = a by a, the horizontal line y = b by b, and the line y = cx through the origin, by c. Then with similar definitions for TB and TC we have

 a' = TA(a) = 1/a, b' = TB(b) = 1/b, and c' = TC(c)

Further
 C' = a'b = (1/a, b), B' = c'a = (a, a/c), and A' = b'c = (1/(bc), 1/b).

Thus AA' has equation x = 1/(bc), BB' has equation y = a/c, and CC' has equation y = (ab)x. Then, AA'BB' = (1/(bc), a/c) which lies on the line CC'.

II.17)Điểm Parry reflection.
Kết quả:: Cho tam giác . Kẻ qua các đường thẳng song song với nhau và song song với đường thẳng Euler của tam giác. Gọi lần lượt là các đường đối xứng với qua . Khi đó các đường này đồng quy tại điểm Parry reflection của tam giác ABC.
Chỉ dẫn chứng minh:

Gọi là trực tâm tam giác , là ảnh của qua phép đối xứng trục . Phép vị tự tâm tỉ số biến tam giác thành tam giác khi đó 2 đường thẳng của 2 tam giác này trùng nhau.

Gọi là giao điểm của với
Ta có:
Vậy cũng là ảnh của qua phép đối xứng trục
Dựa vào tính chất đồng dạng ta suy ra được đường thẳng đối xứng với đường thẳng qua trục
Tương tự với các đường thẳng còn lại ta suy ra 3 đường thẳng đồng quy theo định lí
(Xem them File FG200806.bdf)

II.18)Đường tròn Taylor ,tâm Taylor
Kết quả:.Cho tam giác các đường cao . Từ kẻ các đường vuông góc với các cạnh khi đó chân các đường vuông góc này nằm trên cùng 1 đường tròn gọi là đường tròn của tam giác
Chỉ dẫn chứng minh:

Ta có:

Vậy 4 điểm cùng nằm trên 1 đường tròn.
Tương tự với 2 bộ 4 điểm .
Ta được:
Vậy 4 điểm cũng cùng nằm trên 1 đường tròn hay ta suy ra được dpcm

II.19)Điểm Bevan
Kết quả:. Cho tam giác , là tâm các đường tròn bàng tiếp. Khi đó tâm đường tròn ngoại tiếp tam giác được gọi là điểm của tam giác .

Sau đây ta sẽ đến với 1 vài tính chất cua điểm :

*Ta thấy rằng là tâm đường tròn nội tiếp tam giác chính là trực tâm của tam giác khi đó đường tròn ngoại tiếp tam giác ABC là đường tròn chín điểm của tam giác suy ra là trung điểm của
*Tâm là trung điểm của đoạn nối trực tâm với điểm .
Đầu tiên ta chứng minh 3 điểm này thẳng hàng.
Xét tam giác ta có:

Vậy 3 điểm này thẳng hàng.
Xét tam giác ta có 3 điểm thẳng hàng suy ra:

Vậy ta có dpcm
*Điểm là trung điểm của đoạn nối điểm và điểm .
Điểm là điểm đối xứng của trực tâm qua tâm đường tròn ngoại tiếp.
Khi đó tương tự như trên ta cũng có được dpcm

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