Two inequalities about the pedal triangle
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Abstract
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.
Keywords
Interior point Pedal triangle Inequality Automated inequality provingMSC
51M16 12041 Introduction
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.
2 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 lefthand side inequality of (4). For the righthand 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.
2.1 An upper bound of \(R_{1}R_{2}R_{3}\)
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])
In this lemma, \(\boldsymbol {E}_{n}\) denotes the real ndimensional Euclidean space and \(\\boldsymbol {x}\\) is the Euclidean norm of x.
 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 \(\triangle ABC\).
 2.
The circumcenter O is inside \(\triangle ABC\) and P coincides with O.
 3.
P coincides with one of the vertices of \(\triangle ABC\).
2.2 A lower bound of \(R_{1}R_{2}R_{3}\)
For the righthand side of (4), we have the following.
Proposition 1
Proof
We discuss this problem in three cases according to the position of O.
I.O is outside \(\triangle ABC\).

\(P_{1}P_{2}\) is a diameter of the circumcircle.

\(P_{1}\) is on the minor arc $\stackrel{\u23dc}{{B}_{2}C{A}_{2}}$.

The minor arc $\stackrel{\u23dc}{{P}_{1}{A}_{2}}$ is smaller than the arc $\stackrel{\u23dc}{{P}_{1}{A}_{2}A}$.

The minor arc $\stackrel{\u23dc}{{P}_{1}{B}_{2}}$ is smaller than the arc $\stackrel{\u23dc}{{P}_{1}{B}_{2}B}$.
II.O is on a side of \(\triangle ABC\).
Assume that O is on the side AB of \(\triangle ABC\). Draw a line passing through P and parallel to AB. By a similar way, we can prove (8) for any interior point P of \(\triangle ABC\).
III.O is inside \(\triangle ABC\).
In this case, we need partition \(\triangle ABC\) 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.

\(A_{2}\) lies on the minor arc $\stackrel{\u23dc}{C{A}_{0}}$ and \(B_{2}\) lies on the minor arc $\stackrel{\u23dc}{{B}_{0}C}$.

\(P_{2}\) lies on the minor arc $\stackrel{\u23dc}{AB}$ and \(P_{1} P_{2}\) is a diameter of the circumcircle.

\(P_{1}\) lies on the minor arc $\stackrel{\u23dc}{{B}_{2}C{A}_{2}}$.

\(\angle P_{1} OA_{2}<\angle P_{1} OA_{0}<\angle P_{1} OA\) and \(\angle P_{1} OB_{2}<\angle P_{1} OB_{0}<\angle P_{1} OB\).
From all above, we achieve that (8) holds for every interior point P of \(\triangle ABC\) 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 \(\triangle ABC\) and the equality takes place if and only if P coincides with the circumcenter.
3 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.
Lemma 2
Proof
We use the distance geometry method to achieve this equation.
From all above, (31) is proved, so are (25) and (24). That is to say, (2) is correct for \(\triangle ABC\) and its interior point P, and the equality of (2) holds if and only if \(\triangle ABC\) is an equilateral triangle and P is its circumcenter.
Remark 1
Since \(h_{0,0}\geq0\) when p, q are both positive, \(h_{0,0}_{p=0}=1\), \(h_{0,0}_{q=0}=(p+1)^{12}\), there must exist \(\partial h_{0,0}/\partial p=0 \wedge\partial h_{0,0}/\partial q=0\), if the equality \(h_{0,0}=0\) holds. Then we could use the Maple function RealRootCounting to show that there is only one real solution for the semialgebraic system \(\{h_{0,0}=0, \partial h_{0,0}/\partial p=0, \partial h_{0,0}/\partial q=0, p>0, q>0\}\). Because \(h_{0,0}_{p=2, q=3/2}=0\), it just presents a proof to the equivalent (32).
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 \(f\geq0\) 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 I53230 CPU and 8GB RAM. This package is available at http://faculty.uestc.edu.cn/huangfangjian/en/article/167349/content/2378.htm.
Remark 3
 1.when \(a=10\), \(b=2\), \(c=9\), \(x=1/10\), \(y=1/10\), \(z=4/5\), we have$$\begin{aligned}& R_{1}R_{2}R_{3}\approx15,\qquad \frac{r^{2}(R^{2}PO^{2})^{2}}{2r_{1}r_{2}r_{3}} \approx130,\qquad (RPO) (R+PO)^{2}\approx92, \end{aligned}$$(33)
 2.when \(a=10\), \(b=6\), \(c=8\), \(x=1/2\), \(y=1/3\), \(z=1/6\), we have$$\begin{aligned}& R_{1}R_{2}R_{3}\approx88,\qquad \frac{r^{2}(R^{2}PO^{2})^{2}}{2r_{1}r_{2}r_{3}} \approx115, \qquad(RPO) (R+PO)^{2}\approx142. \end{aligned}$$(34)
4 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 computeraided 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.
Notes
Acknowledgements
The figures in this paper are drawn on the website http://www.netpad.net.cn. It is an efficient mathematical drawing tool especially for the geometric objects. The author gratefully appreciates the convenience of this platform. Additionally, the author acknowledges the support provided by the National Natural Science Foundation of China (grant No. 61374001) and appreciates the reviewers for their careful reading and valuable suggestions.
Authors’ contributions
The author read and approved the final manuscript.
Competing interests
The author declares that he does not have any commercial or associative interest that represents a conflict of interest in connection with the work submitted.
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