Divide ABCDE into two parts with equal area

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SUMMARY

The discussion focuses on dividing a convex pentagon ABCDE into two equal-area parts by constructing a line through point A. For a regular pentagon, the midpoint of segment CD, denoted as F, can be used to draw line AF, effectively splitting the pentagon. The conversation references Albert's solution, which involves finding the midpoint M of segment PQ and drawing line AM, contingent on M being positioned between points C and D. However, the participants highlight that this condition may not always hold true, indicating the need for further exploration of the geometric properties involved.

PREREQUISITES
  • Understanding of convex polygons, specifically pentagons.
  • Knowledge of geometric constructions, including midpoints and line segments.
  • Familiarity with area calculations in polygons.
  • Basic principles of geometric proofs and visual representations.
NEXT STEPS
  • Study geometric constructions involving midpoints in polygons.
  • Learn about area division techniques in irregular shapes.
  • Explore geometric proof strategies, including "proof without words".
  • Investigate the properties of convex polygons and their area relationships.
USEFUL FOR

Mathematicians, geometry students, and educators interested in polygon properties and area division techniques will benefit from this discussion.

Albert1
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ABCDE is a pentagon,please construct a line (passing

through point A),and divide ABCDE into two parts with equal

area
 
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Is this a general pentagon, or a particular pentagon?
 
For a regular pentagon, just construct the midpoint of $CD$, call if $F$, and draw the segment $AF$. This splits the pentagon in two equal pieces.
 
Prove It said:
Is this a general pentagon, or a particular pentagon?
A general pentagon(convex)
 
Albert said:
ABCDE is a pentagon,please construct a line (passing

through point A),and divide ABCDE into two parts with equal

area
http://www.mathhelpboards.com/attachments/f28/998d1373594278-change-pentagon-into-triangle-equal-area-pentagon.jpg

​
Referring to Albert's beautiful solution to the problem in http://www.mathhelpboards.com/f28/change-pentagon-into-triangle-equal-area-5486/, if $M$ is the midpoint of $PQ$ then the line $AM$ will do the job, provided that $M$ lies between $C$ and $D$. I imagine that this must necessarily be the case, but I don't see how to prove it.
 
In my opinion the best proof is "a proof without words"
so again I construct a diagram and let it explain the solution
View attachment 1033
and AG is what we need as written by Opalg
"if [FONT=MathJax_Math]M is the midpoint of [FONT=MathJax_Math]P[FONT=MathJax_Math]Q then the line [FONT=MathJax_Math]A[FONT=MathJax_Math]M will do the job"
 

Attachments

  • pentagon02.jpg
    pentagon02.jpg
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Last edited:
Opalg said:
Referring to Albert's beautiful solution to the problem in http://www.mathhelpboards.com/f28/change-pentagon-into-triangle-equal-area-5486/, if $M$ is the midpoint of $PQ$ then the line $AM$ will do the job, provided that $M$ lies between $C$ and $D$. I imagine that this must necessarily be the case, but I don't see how to prove it.
The statement M lies between C and D is not always true
in fact M and C (or M and D)may coincide
May M also lie between D and E ?(if the length of CD is very small)
M may also lie between B and C.
(we may check this using various diagram)
 
Last edited:

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