A given convex pentagon $ ABCDE$ has the property that the area of each of five triangles $ ABC, BCD, CDE, DEA$, and $ EAB$ is unity (equal to 1). Show that all pentagons with the above property have the same area, and calculate that area. Show, furthermore, that there are infinitely many non-congruent pentagons having the above area property.
Problem
Source: 1972 USAMO Problem 5
Tags: geometry, parallelogram
futuremospper
06.03.2010 06:13
Since $ [EDC]=[BDC]=1$, both triangles have equal altitudes to $ \overline{CD}$. Hence $ \overline{DC} \parallel \overline{EB}$. Similarly, the other diagonals are parallel to their respective opposite sides. Hence, $ ABPE$ is a parallelogram and $ [PEB]=1$. Letting $ [EDP]=y=[PBC]$ and $ [PDC]=x$, we then have $ x+y=1$. Also:
\[ \frac{[EPD]}{[EPB]} = \frac{DP}{PB}=\frac{[DPC]}{[CPB]}\]
or
\[ \frac{y}{1} = \frac{x}{y},\]
so $ y^2+y-1=0$ and $ y = \frac{\sqrt{5}-1}{2}$. Then $ [ABCDE] = 2-y-x-y = \frac{5+\sqrt{5}}{2}$.
To prove that there is an infinite number of such noncongruent pentagons, construct an arbitrary triangle $ PDC$ whose area is $ x$. Extend $ CP$ to $ E$ and $ DP$ to $ B$, so that $ [EDC]=[BDC]=1$. Now draw $ \overline{EA} \parallel \overline{BD}$ and $ \overline{AB} \parallel \overline{EC}$. It follows from the previous analysis that the pentagon has the desired property.
Vo Duc Dien
29.06.2011 00:32
Interesting! The Bangladeshians (people from Bangladesh) used similar problem for their Mathematical Olympiad 2012 competition: "In a given pentagon ABCDE, triangles ABC, BCD, CDE, DEA and EAB all have the same area. The lines AC and AD intersect BE at points M and N. Prove that BM = EN." Let me suggest these similar problems: 1. Given a convex pentagon ABCDE with M, N, R, S and T be the intersections of AC and BE, AD and BE, AD and CE, BD and CE, BD and AC, respectively such that the areas of triangles ABM, BCT, CDS, DER and AEN are also the same. Prove that ABCDE is a regular pentagon. 2. Given a convex pentagon ABCDE, the areas of triangles ABC, BCD, CDE, DEA and ABE are all the same. Let M, N, R, S and T be the intersections of AC and BE, AD and BE, AD and CE, BD and CE, BD and AC, respectively. Prove that the areas of triangles ABM, BCT, CDS, DER and AEN are also the same.
wxl18
03.12.2022 09:16
redacted
mannshah1211
03.12.2022 10:20
wxl18 wrote:
Let $P(A)$ and $P(B)$ be the probability that the product does not include an even number and the probability that the product does not include the number $5$, respectively. Notice that in order for the final product to be divisible by $10$, it must include an even number and the number $5$. Thus, we want the complement of $P(A\cup B)$. By PIE, $1-P(A\cup B)=1-(P(A)+P(A)-P(A\cap B)=1-((\frac{8}{9})^n+(\frac{5}{9})^n-(\frac{4}{9})^n).\blacksquare$
I think you may have posted on the wrong topic.
huashiliao2020
13.04.2023 05:18
Vo Duc Dien wrote: Interesting! The Bangladeshians (people from Bangladesh) used similar problem for their Mathematical Olympiad 2012 competition: "In a given pentagon ABCDE, triangles ABC, BCD, CDE, DEA and EAB all have the same area. The lines AC and AD intersect BE at points M and N. Prove that BM = EN." Let me suggest these similar problems: 1. Given a convex pentagon ABCDE with M, N, R, S and T be the intersections of AC and BE, AD and BE, AD and CE, BD and CE, BD and AC, respectively such that the areas of triangles ABM, BCT, CDS, DER and AEN are also the same. Prove that ABCDE is a regular pentagon. 2. Given a convex pentagon ABCDE, the areas of triangles ABC, BCD, CDE, DEA and ABE are all the same. Let M, N, R, S and T be the intersections of AC and BE, AD and BE, AD and CE, BD and CE, BD and AC, respectively. Prove that the areas of triangles ABM, BCT, CDS, DER and AEN are also the same. Cool, are you the creator of these problems? How did you post used a similar problem for 2012 Olympiad when your post was in 2011!? Anyways, back to the problem. futuremospper wrote:
Since $ [EDC]=[BDC]=1$, both triangles have equal altitudes to $ \overline{CD}$. Hence $ \overline{DC} \parallel \overline{EB}$. Similarly, the other diagonals are parallel to their respective opposite sides. Hence, $ ABPE$ is a parallelogram and $ [PEB]=1$. Letting $ [EDP]=y=[PBC]$ and $ [PDC]=x$, we then have $ x+y=1$. Also:
\[ \frac{[EPD]}{[EPB]} = \frac{DP}{PB}=\frac{[DPC]}{[CPB]}\]
or
\[ \frac{y}{1} = \frac{x}{y},\]
so $ y^2+y-1=0$ and $ y = \frac{\sqrt{5}-1}{2}$. Then $ [ABCDE] = 2-y-x-y = \frac{5+\sqrt{5}}{2}$.
To prove that there is an infinite number of such noncongruent pentagons, construct an arbitrary triangle $ PDC$ whose area is $ x$. Extend $ CP$ to $ E$ and $ DP$ to $ B$, so that $ [EDC]=[BDC]=1$. Now draw $ \overline{EA} \parallel \overline{BD}$ and $ \overline{AB} \parallel \overline{EC}$. It follows from the previous analysis that the pentagon has the desired property.
sorry, what was point P?