What Does PB Refer to in Measurements?

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Discussion Overview

The discussion revolves around determining the length of segment PB in a geometric configuration involving a square and points A, B, O, and P. The context includes mathematical reasoning and the application of geometric principles such as rotation and congruence of triangles.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • Some participants suggest rotating the square by 90 degrees clockwise to analyze the position of point P relative to points O and B.
  • Others propose an alternative approach by rotating the square 90 degrees anticlockwise, asserting that the new position of P (denoted as P') will lie on line PB and that triangles APB and BP'C are congruent.
  • It is noted that if the lengths of segments AP and BP are defined as 5x and 14x respectively, then applying the Pythagorean theorem leads to the conclusion that the area of the square is 221x².
  • Participants calculate that since 1989 equals 9 times 221, it follows that x equals 3, leading to the conclusion that the length of PB is 42 cm.

Areas of Agreement / Disagreement

There appears to be some agreement on the calculations leading to the length of PB, but the methods of rotation and the implications of congruence are discussed with different approaches, indicating a lack of consensus on the best method.

Contextual Notes

The discussion relies on specific geometric assumptions and the congruence of triangles, which may not be universally accepted without further clarification of the configuration.

Albert1
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please find the length of PB

View attachment 1178
 

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Albert said:
please find the length of PB

View attachment 1178
HINT:

Rotating the square by 90 degrees in the clockwise direction, the new position of $P$ is collinear with $P$ and $O$.
 
http://mathhelpboards.com/attachments/challenge-questions-puzzles-28/1178d1376578218-find-length-pb-length-pb.jpg
caffeinemachine said:
HINT:

Rotating the square by 90 degrees in the clockwise direction, the new position of $P$ is collinear with $P$ and $O$.
Good idea! But what I would do is to rotate the square by 90 degrees in an anticlockwise direction. The new position (call it $P'$) of $P$ will be on the line $PB$. The lines $PB$, $P'C$ will be perpendicular, and the triangles $APB$, $BP'C$ will be congruent. Therefore $\angle APB$ is a right angle. If the lengths of $AP$, $BP$ are $5x$ and $14x$ then by Pythagoras $AB$ will be $\sqrt{221}x$ and the area of the square is $221x^2$. Since $1989 = 9\times 221$ it follows that $x=3$ and so $PB = 42$cm.
 
It is easy to see that four points A,B,O,P are
located on the same circle
let AP=5x ,BP=14X
$AP^2+BP^2=221x^2=AB^2=1989$
$\therefore x=3cm$
$BP=14x=42cm$
View attachment 1184
 

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Last edited:

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