Two inequalities about the pedal triangle

Two conjectures about the pedal triangle are proved. For the first conjecture, the product of the distances from an interior point to the vertices is mainly considered and a lower bound is obtained by the geometric method. To prove the other one, an analytic expression of the distance between the circumcenter and an interior point is achieved by the distance geometry method. A procedure to transform the geometric inequality to an algebraic one is presented. And then the proof is finished with the help of a Maple package, Bottema. The proof process could be applied to similar problems.


Introduction
For an interior point P of a triangle ABC, let D, E, F denote the feet of the perpendiculars from P to BC, CA, AB (may be produced), respectively. Then the triangle DEF is the pedal triangle of P with respect to ABC shown in Fig. 1. It is an elementary geometric object and has been introduced in many textbooks such as [1] and [2], in which lots of theorems about the pedal triangle were presented. Most of these results are equalities. In [3], Liu puts forward some inequalities involving pedal triangles.
Let O, R, r, S denote the circumcenter, circumradius, inradius, and the area of ABC, respectively; a, b, c denote the lengths of line segments BC, CA, AB; R 1 , R 2 , R 3 , r 1 , r 2 , r 3 denote the distances from P to A, B, C, D, E, F, respectively; and R p denotes the circumradius of DEF , shown in Fig. 2.
In the last section of [3], some conjectures were presented. For conjectures (3.4) and (3.5), we determine that they are both correct. Notations as above, these two conjectures are as follows: and Here, PO is the distance from P to the circumcenter O. (Actually, |PO| is more formal, however, we usually use PO when there is no confusion.) Although these two conjectures are about R p , the circumradius of DEF, by the following well-known equation (Theorem 198, Corollary C, [2]), we get an equivalent inequality of (1) There are only three geometric variables involved above. One is R 1 R 2 R 3 , the product of the distances from P to A, B, C, the other two are the circumradius of ABC and the distance from P to O. We could prove it in a geometric way. For conjecture (2), an equivalent inequality by (3) is However, there are more variables in this inequality. We prove it in an algebraic way instead of a geometric one. The remaining parts are arranged as follows. First, according to the position of circumcenter, we prove conjecture (4) in three subcases in Sect. 2. After that, an analytic expression of PO is obtained by the distance geometry method [4]. Based on this expression, conjecture (2) is transformed and proved with the help of a Maple package Bottema [5] in Sect. 3. We also compare these two upper bounds of R 1 R 2 R 3 in the last part.

Proof to the first conjecture
In this section, we present a geometric proof to conjecture (1). First, we recall a result, Theorem 2 of [6] for the left-hand side inequality of (4). For the right-hand side, we divide it into three subcases to construct this lower bound of R 1 R 2 R 3 according to the position of O in Proposition 1.

An upper bound of
For a point in a polytope, [6] presented an upper bound of the product of the distances from the vertices. We just list them here.
Lemma 1 (Theorem 2 in [6]) Let x, x 1 , . . . , x m (m ≥ 2) be (not necessarily distinct) points of the solid unit sphere U n of E n such that x belongs to the convex hull of For 0 < x < 1, equality holds in (6) only under the following conditions: In this lemma, E n denotes the real n-dimensional Euclidean space and x is the Euclidean norm of x.
When considering a triangle ABC and a point P that belongs to it, based on this lemma, we have and the equality holds if and only if one of the following conditions holds: 1. P lies on the chord joining two points of {A, B, C}, and the remaining one is farthest away from P on the circumcircle of ABC. 2. The circumcenter O is inside ABC and P coincides with O. 3. P coincides with one of the vertices of ABC. That is to say, when P is an interior point of ABC, we have inequality (7) and the equality holds only when P coincides with O.

A lower bound of R 1 R 2 R 3
For the right-hand side of (4), we have the following.

Proposition 1 Notations as above, for any interior point P of ABC, we have
and the equality holds only when P coincides with the circumcenter O.
Proof We discuss this problem in three cases according to the position of O.
In this case, there must exist one side of ABC (e.g., AB) such that O and the remaining point (C) are located on its different sides (see Fig. 3).
For any interior point P of ABC, draw a line passing through P, parallel to AB and meeting the circumcircle in two points, then one point must be on the minor arc AC and the other must be on the minor arc CB. Let A 2 denote the first one and the latter is denoted by B 2 . Produce OP to intersect the circumcircle at two points P 1 and P 2 in which P 1 is on the minor arc ACB and P 2 is on the major arc AB. Details are shown in Fig. 4. Then we have: • P 1 P 2 is a diameter of the circumcircle.
• P 1 is on the minor arc B 2 CA 2 .
• The minor arc P 1 A 2 is smaller than the arc P 1 A 2 A.
• The minor arc P 1 B 2 is smaller than the arc P 1 B 2 B. Therefore, ∠POA 2 = ∠P 1 OA 2 < ∠P 1 OA = ∠POA and ∠POB 2 = ∠P 1 OB 2 < ∠P 1 OB = ∠POB. Let us compare POA 2 and POA. There is a common side PO. OA 2 and OA are both R, the circumradius. According to the law of cosines, we have PA > PA 2 . Similarly, we have PB > PB 2 . Since the chord A 2 B 2 and the diameter P 2 P 1 intersect at the point P, according to the intersecting chords theorem (also known as power of a point or secant tangent theorem), we have PA 2 · PB 2 = PP 1 · PP 2 = (R -PO) · (R + PO). Additionally, we have PC ≥ |OC -OP| = R -PO and the equality holds only when P lies on the radius OC. Then there exist Assume that O is on the side AB of ABC. Draw a line passing through P and parallel to AB. By a similar way, we can prove (8) for any interior point P of ABC.
III. O is inside ABC.
In this case, we need partition ABC to three quadrilaterals. Produce AO, BO, CO to meet BC, CA, AB in A 1 , B 1 , C 1 respectively, so P must lie inside one of the quadrilaterals CB 1 OA 1 C, B 1 AC 1 OB 1 and C 1 BA 1 OC 1 , or on OA 1 or OB 1 or OC 1 .
When P coincides with O, PO = 0 and R 1 = R 2 = R 3 = R, then the equality of (8) holds.

Figure 5 O is inside. Auxiliary lines when P is in the quadrilateral
When P lies in a quadrilateral, say CB 1 OA 1 C (see Fig. 5), let us draw a line which is parallel to AB, passes through P and meets the circumcircle in two points in which one is on the minor arc CA and the other is on the minor arc BC. Let A 2 and B 2 denote the former and the latter, respectively. Draw a line from O to P and produce it to intersect the circumcircle at P 1 . Draw another line from P to O and produce it to meet the circumcircle again in P 2 . And draw a line passing through O, parallel to AB and meeting the circumcircle in two points. Let A 0 (B 0 ) denote the point on the minor arc CA ( BC). Therefore, the following properties are easy to prove: • A 2 lies on the minor arc CA 0 and B 2 lies on the minor arc B 0 C.
• P 2 lies on the minor arc AB and P 1 P 2 is a diameter of the circumcircle.
• P 1 lies on the minor arc B 2 CA 2 .
• ∠P 1 OA 2 < ∠P 1 OA 0 < ∠P 1 OA and ∠P 1 OB 2 < ∠P 1 OB 0 < ∠P 1 OB. Once again, comparing POA 2 and POA, there is a common side PO and OA, OA 2 are both circumradius, then we have PA > PA 2 according to the law of cosines. Similarly, we could have PB > PB 2 . Applying the intersecting chords theorem to the chord A 2 B 2 and the diameter P 1 P 2 , we have Additionally, PC ≥ OC -PO = R -PO and the equality holds only when P lies on the radius OC of the circumcircle, we have When P is in the quadrilateral B 1 AC 1 OB 1 (C 1 BA 1 OC 1 ), we draw the parallel line of BC (AC) through P. When P is on the line segment OA 1 (OB 1 , OC 1 ) and does not coincide with O, we draw the parallel line of AB (BC, AC) through P, respectively. And in a similar way, we could obtain (8).
From all above, we achieve that (8) holds for every interior point P of ABC and the equality holds only when P coincides with O.
Based on (7), (8), and (4), we determine that (1) is correct for any interior point P of ABC and the equality takes place if and only if P coincides with the circumcenter.

Proof to the second conjecture
First we use the barycentric coordinate system and the distance geometry method to present an analytic expression of PO. And then we transform conjecture (2) equivalently to a polynomial inequality with four variables. After that, an inequality proving tool, Maple package Bottema developed by Prof. Lu Yang and his collaborators, is invoked to help us prove it.
Let (x, y, z) denote the barycentric coordinates of the interior point P with respect to ABC. And we choose the normalized coordinates. That is to say, in which S PBC denotes the area of the triangle PBC, similar as S PCA and S PAB do. Therefore, we have x + y + z = 1. There are also some well-known formulas, we just list them below without proof.
What we need more is an explicit expression of PO.

Lemma 2 For any interior point P of ABC, notations as above, we have
Proof We use the distance geometry method to achieve this equation.
Let ABC be the reference triangle, for any point Q on the plane of ABC, we take Since these five points are on the same 2-dimensional plane, the (4, 5) minor of CM vanishes [7], i.e., By solving this equation, we obtain Based on (13) and (14), we have Consequently, we achieve (18).
Remark 2 The function xprove in the Maple package Bottema is based on the dimensionaldecreasing algorithm ( [5], Chap. 8) and the complete discrimination system for polynomials [8]. It is quite a powerful tool for automated inequality proving; however, due to the expansion of symbolic computation, when there are too many variables and the degree is too high, the calculation will not be very efficient. The direct proof by xprove to f ≥ 0 is not practical. We tried for more than six hours, but nothing returned. In our proof, it takes three minutes to transform and prove the inequalities in Maple 2016 on a laptop with Intel I5-3230 CPU and 8GB RAM. This package is available at http://faculty.uestc.edu.cn/huangfangjian/en/article/167349/content/2378.htm. (4) and (5) could both be treated as the upper bounds of R 1 R 2 R 3 , the product of the distances from an interior point to the vertices of a triangle. Actually, we once tried to find the larger one between them, however, examples show that the comparison result varies. For example, 1. when a = 10, b = 2, c = 9, x = 1/10, y = 1/10, z = 4/5, we have

Conclusion
In this paper, we have proved two interesting conjectures about the pedal triangle of an interior point of a triangle and analyzed the conditions when the equalities hold. We present a geometric method to deal with the first one. For the second one, we use some algebraic equations to transform it to a polynomial inequality and divide it into some inequalities with fewer variables and lower degrees. And then a computer-aided tool is invoked to finish the proof. As we know, there are plenty of inequality proving algorithms and methods. Taking advantages of these tools, we could think about complex issues. The procedure of the latter proof could be applied to other similar problems.