Let $ABC$ be a triangle with incentre $I$. A point $P$ in the interior of the triangle
I will only give a brief explanation to the solution of this problem.
Referring to the diagram below,
we need the following knowledge:- Let I be the in-center of $\triangle ABC$. The perpendicular bisector of BC and the angle bisector of $\angle A$ will meet at X and X is right on circumference of the circle ABC. In addition, X happens to be the center of the circle passing through B, I, C. Then, XB = XI = XC.
Based on the given, after some calculation, we get:-
∠IBP = ∠IBC - ∠PBC = 1/2 ∠ABC - ∠PBC = 1/2 [∠PBA - ∠PBC] …. (1)
Similarly, ∠ICP = ∠PCB - ∠ICB = ∠PCB - 1/2 ∠ACB = 1/2[∠PCB - ∠PCA] …. (2)
Since ∠PBA +∠PCA = ∠PBC +∠PCB , then ∠PBA - ∠PBC = ∠PCB - ∠PCA …. (3)
(1) , (2) and (3) imply: ∠IBP = ∠ICP and therefore BIPC is cyclic. This is equivalent to adding P as another con-cyclic point to the circle BIC. See the diagram below,
In $\triangle PAX$, by triangle inequality, we have $AP + PX \ge AX = AI + IX$
Result follows after subtracting $PX = IX$ from both sides.
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pi-π
Updated on December 09, 2022Comments
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pi-π 11 months
Let $ABC$ be a triangle with incentre $I$. A point $P$ in the interior of the triangle satisfies $\angle PBA+\angle PCA=\angle PBC+\angle PCB$. Show that $AP\geq AI$, and the equality holds if and only if $P=I$.
Please help me. I couldn't get anything from the question..
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pi-π almost 7 years@@J.H..Heller, Could you please provide me with a figure?
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J. Heller almost 7 yearsThis is actually a very hard question. Based on looking at some of your more recent questions (for example, chord of length 16 in radius 10 circle), this question is a little beyond your level right now. Once you've mastered more plane geometry (circumcircles, inscribed circles, inscribed angles, cyclic quadrilaterals), you might want to come back to this question. Then it would only take a few minutes to draw your own figure on paper to follow along with the solution.
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pi-π almost 7 years@@J.Heller However, I know all those concepts (cyclic quadrilateral, inscribed angles, inscribed circles, etc).. But, I.just couldnot draw a fig. so, If you please provide me a figure, it would be helpful. Thanks..
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J. Heller almost 7 yearsThere is a difference between "knowing" and "mastering". Someone with a very good grasp of all these concepts could easily draw a figure of this problem on paper in a few minutes. It would take me longer than this to create a good quality electronic figure.
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pi-π almost 7 years@ Mick,. and how about $P=I$?
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Mick almost 7 years@NeWtoN P = I means P is right on I. From the diagram, one can see that the blue triangle is degenerated to a straight line only. Then, AP = AI.
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pi-π almost 7 yearsDo we need both diagrams?
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Mick almost 7 years@NeWtoN Two diagrams make the explanation easier to understand.
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J. Heller almost 7 years@Mick: The circumcircle of ABC is a distraction, it is completely unnecessary for solving the problem. You say the circumcenter of $CBI$ "just happens" to be on the line thru $AI.$ Are you aware of some theorem that says that the angle bisector of a triangle vertex passes thru the circumcenter of the triangle defined by incenter and opposite edge? It must be fairly obscure because I couldn't find any such theorem on Wikipedia (looking in subjects such as incenter, circumcenter, angle bisector).
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Mick almost 7 years@J.Heller Let’s start from the circum-circle ABC. The blue line is the perpendicular bisector of BC. It will cut the circle at some point X. Then, $\angle BCX = \angle CBX$. This further means arc(BX) = arc(CX). Next, we join AX. (At this point, we are not sure whether AX will pass through I or not.) Then, $\angle BAX = \angle CAX$ because they are subtending an equal arc. From that, we conclude that AX is actually the angle bisector of $\angle BAC$.
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J. Heller almost 7 years@Mick I'll repeat: you don't need the circumcircle of $\triangle ABC$. What you do need to do is to show that the line thru $AI$ intersects the circumcenter of $\triangle CIB$. In your comment, you did not show that $X$ is the circumcenter of $\triangle CIB$.
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Mick almost 7 years@J.Heller Continued. In addition to XB = XC, we also have XI = XC because both $\angle XIC$ and $\angle XCI $ are equal to $0.5\angle BAC + 0.5\angle ACB$ (via $\angle BAX$). This makes X the circum-cneter of $\triangle CIB$.