The longest diagonal of a convex hexagon is $2$. Is there necessarily a side or diagonal in this hexagon whose length does not exceed $1$?
Problem
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Tags: geometry, geometric inequality
Spakian
16.07.2021 01:44
Let us assume the hexagon in the attachment is regular. Going off of the attached image, we see that $AD = 2$ because it's the longest diagonal. We know that $\angle ADE = 60$ degrees and $\angle EAD = 30$ degrees because $ABCDEF$ is regular (so the interior angle length is $120$ degrees). We have a $30-60-90$ triangle with hypotenuse 2 so that means $ED = 1$ and $AE = \sqrt{3} > 1$. Since $AE$ is a diagonal and $\sqrt{3} > 1$, there is a diagonal whose length exceeds 1
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StopSine
16.07.2021 01:55
Spakian wrote:
Let us assume the hexagon in the attachment is regular. Going off of the attached image, we see that $AD = 2$ because it's the longest diagonal. We know that $\angle ADE = 60$ degrees and $\angle EAD = 30$ degrees because $ABCDEF$ is regular (so the interior angle length is $120$ degrees). We have a $30-60-90$ triangle with hypotenuse 2 so that means $ED = 1$ and $AE = \sqrt{3} > 1$. Since $AE$ is a diagonal and $\sqrt{3} > 1$, there is a diagonal whose length exceeds 1
That is not what the question is asking. The question is asking that if the longest diagonal of a convex hexagon is $2$, must there be a side or diagonal whose length is $\leq 1$. In this problem, finding a simple counter-example, that every diagonal/side $>1$, would suffice, but so would showing that such a counter-example doesn't exist.
eagles2018
16.07.2021 02:58
@stopsine That is not what the question is asking either. We can rephrase the question as such: Suppose we have the (infinite) set of all hexagons whose longest diagonal is 2. Does there exist at least one hexagon in this set which has every side and diagonal greater than or equal to 1?
Spakian
16.07.2021 03:49
@above I think the question is asking "that for all convex hexagons whose longest diagonal is 2, is there a side/diagonal whose length is less than 1?"
Spakian
16.07.2021 03:51
Spakian wrote:
Let us assume the hexagon in the attachment is regular. Going off of the attached image, we see that $AD = 2$ because it's the longest diagonal. We know that $\angle ADE = 60$ degrees and $\angle EAD = 30$ degrees because $ABCDEF$ is regular (so the interior angle length is $120$ degrees). We have a $30-60-90$ triangle with hypotenuse 2 so that means $ED = 1$ and $AE = \sqrt{3} > 1$. Since $AE$ is a diagonal and $\sqrt{3} > 1$, there is a diagonal whose length exceeds 1
Basing this on my own idea of what the question asks, there is a side whose length does not exceed $1$. Since the triangle is $30-60-90$, $ED = 1$ and since $ABCDEF$ is regular, all side lengths are equal to 1. Thus, there is a side whose length does not exceed 1
greenturtle3141
16.07.2021 05:40
Oh please. Let me make this crystal clear: - If your answer to the question is yes, then you must prove that ANY convex hexagon with longest diagonal of length 2 MUST have some side or diagonal of length less than or equal to 1. - If your answer to the question is no, then you must find ONE convex hexagon with longest diagonal of length two such that EVERY side and diagonal has length strictly greater than 1. Spakian wrote:
Let us assume the hexagon in the attachment is regular. Going off of the attached image, we see that $AD = 2$ because it's the longest diagonal. We know that $\angle ADE = 60$ degrees and $\angle EAD = 30$ degrees because $ABCDEF$ is regular (so the interior angle length is $120$ degrees). We have a $30-60-90$ triangle with hypotenuse 2 so that means $ED = 1$ and $AE = \sqrt{3} > 1$. Since $AE$ is a diagonal and $\sqrt{3} > 1$, there is a diagonal whose length exceeds 1
This makes no sense. You've found one convex hexagon of longest diagonal 2 that has a diagonal whose length exceeds 1. This does not contribute to either YES or NO in any way. StopSine wrote: Spakian wrote:
Let us assume the hexagon in the attachment is regular. Going off of the attached image, we see that $AD = 2$ because it's the longest diagonal. We know that $\angle ADE = 60$ degrees and $\angle EAD = 30$ degrees because $ABCDEF$ is regular (so the interior angle length is $120$ degrees). We have a $30-60-90$ triangle with hypotenuse 2 so that means $ED = 1$ and $AE = \sqrt{3} > 1$. Since $AE$ is a diagonal and $\sqrt{3} > 1$, there is a diagonal whose length exceeds 1
That is not what the question is asking. The question is asking that if the longest diagonal of a convex hexagon is $2$, must there be a side or diagonal whose length is $\leq 1$. In this problem, finding a simple counter-example, that every diagonal/side $>1$, would suffice, but so would showing that such a counter-example doesn't exist. This is a correct formulation. eagles2018 wrote: @stopsine That is not what the question is asking either. We can rephrase the question as such: Suppose we have the (infinite) set of all hexagons whose longest diagonal is 2. Does there exist at least one hexagon in this set which has every side and diagonal greater than or equal to 1? You're literally saying the same thing. Except the opposite of "does not exceed" is actually "exceeds" i.e. "is strictly greater than", not "greater than or equal to" as you wrote. Spakian wrote: @above I think the question is asking "that for all convex hexagons whose longest diagonal is 2, is there a side/diagonal whose length is less than 1?" Please tell me what words in the question contribute to this interpretation. parmenides does not post trivialities.
greenturtle3141
16.07.2021 05:58
Actually just in case, I'm going to spell out exactly what this question is saying and why the English words in the question say it. The question is of the form: Quote: We have [thing]. Does [thing] necessarily satisfy [some property]? The key word is "necessarily". If you're not sure what this means, let's take a detour into non-math territory. Quote: I found a dress. Will it necessarily look good on me? This quote is of the same form. It asks a yes or no question. You have to answer yes or no. If you answer YES, then you are claiming that: YES wrote: Your dress will necessarily look good. Rephrasing: YES wrote: Your dress MUST look good. Rephrasing for logic: YES wrote: EVERY dress will look good on you. If instead your answer is NO, then you are claiming: NO wrote: No, the dress you have is not necessarily going to look good. Rephrasing: NO wrote: It is a possibility that your dress won't look good. Rephrasing for logic: NO wrote: There exists a dress that would not look good on you. To recap: - If you claim an answer of YES, then you must prove that every dress you could possibly think of would look good (on the speaker). - If you claim an answer of NO, then you must find or conjure up some horrid-looking dress that would not look good (on the speaker). If you agree with this, then you can apply the same parsing to the given math problem. - Replace "dress" with "a convex hexagon whose longest diagonal has length 2" - Replace "looks good" with "has a side/diagonal whose length is $\leq 1$"
StopSine
16.07.2021 16:08
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