The two regular pentagons share a vertex and an edge. The side of the larger pentagon is twice that of the smaller one. Which triangle has larger area, red or blue? Or do they have the same area?
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2$\begingroup$ Almost identical to this recent hot network question: math.stackexchange.com/questions/5101157/… $\endgroup$Servaes– Servaes2025-10-13 21:44:27 +00:00Commented 5 hours ago
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$\begingroup$ @Servaes. Indeed, the underlying idea is the same: Cavalieri’s principle. $\endgroup$Pranay– Pranay2025-10-13 22:03:32 +00:00Commented 4 hours ago
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$\begingroup$ Nice. And somewhat easier than this pentagon HNQ from a month ago: math.stackexchange.com/q/5095643/207316 $\endgroup$PM 2Ring– PM 2Ring2025-10-14 00:01:50 +00:00Commented 2 hours ago
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1$\begingroup$ @PM2Ring. Yep, that was much harder. I have another pentagon problem which also seems harder at the moment. I’ll post it here or MSE depending on how it goes. $\endgroup$Pranay– Pranay2025-10-14 00:19:52 +00:00Commented 2 hours ago
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1 Answer
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The triangle areas are equal.
Consider any of the triangles. It is easy to see (and check) that we can shift its rightmost vertex parallelly to the opposite triangle side to the vertex of the small pentagon.
The shift keeps the triangle area, so the latter is equal to the area of a triangle whose sides are the side of the small pentagon and its diagonals.
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2$\begingroup$ Thanks for adding the figure. It looks great! $\endgroup$Pranay– Pranay2025-10-13 19:28:38 +00:00Commented 7 hours ago
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1$\begingroup$ @Pranay Sorry for not adding it earlier. I had to work with a paper and then a geometry seminar. $\endgroup$Alex Ravsky– Alex Ravsky2025-10-13 19:34:14 +00:00Commented 7 hours ago
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4$\begingroup$ @MathKeepsMeBusy Basically, it comes down to the formula for calculating the area of a triangle: ½b×h The figure-as-drawn involves moving corners in a manner that preserves the height, and thus the area. $\endgroup$Chronocidal– Chronocidal2025-10-13 19:56:36 +00:00Commented 7 hours ago
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1$\begingroup$ Neat. For me the parallelism of the blue/red line pairs wasn't obvious at first, but it follows from the fact that the diagonal of a pentagon is parallel to its opposite side (and the regularity of the shape in general). $\endgroup$Nuclear Hoagie– Nuclear Hoagie2025-10-13 20:28:29 +00:00Commented 6 hours ago
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5$\begingroup$ @z100. Yep, precisely what I had in mind. It’s worth noting that the statement about the relative sizes of the pentagons is completely irrelevant. It was a red herring. $\endgroup$Pranay– Pranay2025-10-13 20:52:03 +00:00Commented 6 hours ago