- #1
maxkor
- 79
- 0
Inside the triangle $ABC$ is point $P$, such that $BP > AP$ and $BP > CP$. Prove that $\angle ABC$ is acute.
Proving $\angle ABC$ is Acute: Inside the Triangle $ABC$
- Context: MHB
- Thread starter maxkor
- Start date
-
- Tags
- Triangle
Click For Summary
Discussion Overview
The discussion revolves around proving that angle $\angle ABC$ is acute within triangle $ABC$, given a point $P$ inside the triangle with specific distance relationships to the vertices. The focus is on the geometric properties and relationships of angles within the triangle.
Discussion Character
- Exploratory, Technical explanation, Debate/contested
Main Points Raised
- One participant suggests that since $BP > AP$ and $BP > CP$, it follows that $\angle ABC$ must be acute.
- Another participant presents a mathematical argument involving inequalities of angles, concluding that $\angle ABC < 90$ based on the relationships between the angles of the triangle.
- A participant expresses confusion about whether a specific triangle meets the criteria set by the problem.
- Further replies question the validity of the triangle fitting the criteria, indicating a misunderstanding of the problem's requirements.
- A participant acknowledges a misreading of the problem, clarifying that they initially thought the task was to show that triangle $ABC$ itself was acute.
Areas of Agreement / Disagreement
The discussion contains multiple viewpoints, with some participants agreeing on the mathematical approach while others express confusion or challenge the applicability of the criteria to specific triangles. There is no consensus on the proof or the interpretation of the problem.
Contextual Notes
There are unresolved assumptions regarding the specific conditions under which the angle relationships hold, and the discussion reflects varying interpretations of the problem statement.
- #2
mrtwhs
- 47
- 0
We are given that $x>y$ and $w>v$ therefore $x+w > y+v = \angle ABC$
Now $\angle CAB > x$ and $\angle ACB >w$ so $\angle CAB + \angle ACB > x+w > y+v = \angle ABC$
But $\angle CAB + \angle ACB = 180 - \angle ABC$ so $180 - \angle ABC > \angle ABC$ and $\angle ABC < 90$
- #3
- 2,020
- 843
- #4
maxkor
- 79
- 0
- #5
- 2,020
- 843
Aaahh! I was reading the problem wrong. I thought it was asking to show that the triangle ABC was acute. My bad.maxkor said:
-Dan
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