{"id":5307,"date":"2025-09-06T18:12:00","date_gmt":"2025-09-06T22:12:00","guid":{"rendered":"http:\/\/mathfun4kids.com\/mlog\/?p=5307"},"modified":"2025-10-05T23:36:14","modified_gmt":"2025-10-06T03:36:14","slug":"usamts-4-1-36","status":"publish","type":"post","link":"https:\/\/mathfun4kids.com\/mlog\/?p=5307","title":{"rendered":"USAMTS 4\/1\/36"},"content":{"rendered":"\n

During a lecture, each of \"$26$\" mathematicians falls asleep exactly once, and stays asleep for a nonzero amount of time. Each mathematician is awake at the moment the lecture starts, and the moment the lecture finishes. Prove that there are either \"$6$\" mathematicians such that no two are asleep at the same time, or \"$6$\" mathematicians such that there is some point in time during which all \"$6$\" are asleep.\ud83d\udd11<\/a><\/p>\n

<\/p>\n\n\n\n

Proof 1:<\/strong> (Formal)<\/p>\n\n\n\n

Case 1: If there is any group of $6$ or more mathematicians who are asleep at same time at some point during the lecture, the problem is solved.<\/p>\n\n\n\n

Case 2: Assume there is no group of $6$ or more mathematicians who are asleep at same time at some point during the lecture. This implies that:<\/p>\n\n\n\n

(a) There are no more than 5 mathematicians asleep at same time at some point during the lecture.<\/p>\n\n\n\n

(b) Denote each mathematician with an integer number, such as $M_i$, where $1\\le i\\le 26$, based on the order of starting time of their sleeping periods. For example mathematician $M_1$ is the first one who goes asleep, and mathematician $M_{26}$ is the last one who goes asleep.<\/p>\n\n\n\n

(c) Additionally, denote the sleeping period of mathematician $M_i$ as an interval $(a_i, b_i)$, where $a_i$ is the starting time, $b_i$ the ending time, and $a_i<b_i$ ($i=1,2,\\ldots,26$). By (b), we have $$a_1\\le a_2 \\le a_3 \\le \\ldots \\le a_{25} \\le a_{26}$$<\/p>\n\n\n\n

Therefore, for any 6 consecutive starting times of sleeping periods of 6 mathematicians, $M_i, M_{i+1}, M_{i+2}, M_{i+3}, M_{i+4}, M_{i+6}$, their sleep periods will be $${(a_i, b_i), (a_{i+1}, b_{i+1}), (a_{i+2}, b_{i+2}), (a_{i+3}, b_{i+3}), (a_{i+4}, b_{i+4}), (a_{i+5},b_{i+5})}$$<\/p>\n\n\n\n

There exists at least one $(a_j, b_j)$, $i\\le j\\le i+4$, so that it does not overlap with $(a_{i+5}, b_{i+5})$, i.e. $a_j<b_j<a_{i+5}$; otherwise these $6$ sleeping periods all overlap with each other, in contradiction to the assumption (a). This also implies that $(a_j, b_j)$ does not overlap with any sleeping period $(a_k, b_k)$ where $k\\ge i+5$, as $a_j<b_j<a_k<b_k$, which translates to:<\/p>\n\n\n\n

(d) The sleeping period of mathematician $M_j$ does not overlap with the sleeping period of any mathematician $M_k$, where $k\\ge i+5$.<\/p>\n\n\n\n

(e) In this way, we can find $5$ mathematicians, one each from $5$ distinct groups of mathematicians ${M_i, \u2026, M_{i+4}}$ where $i=1,6,11,16,21$, so that the sleeping period of these $5$ mathematicians do not overlap with each other.<\/p>\n\n\n\n

By (d), the sleeping period of mathematician $M_{26}$ does not overlap any of the $5$ mathematicians selected in (e). Therefore, we have found $6$ mathematicians whose sleeping periods do not overlap with each other.<\/p>\n\n\n\n

Combining the above two cases completes the proof.<\/p>\n\n\n\n

Proof 2<\/strong>: (Informal)<\/p>\n\n\n\n

Assume that there are $5$ private rooms, in each room only one person can sleep; and there is a public room that can sleep $21$ people.<\/p>\n\n\n\n

Whenever a mathematician falls asleep, they are instantly teleported to a private room if available, or to the public room if all private rooms are occupied.<\/p>\n\n\n\n

Whenever a mathematician wakes up, they are instantly teleported back to the original place. <\/p>\n\n\n\n

Case 1: If there is a mathematician at any point sleeping in the public room, it means that all private rooms were occupied, so $5$ other mathematicians were asleep in the private rooms. Therefore, there are at least $6$ mathematicians fell asleep at the same time.<\/p>\n\n\n\n

Case 2: If there is no mathematician at any point sleeping in the public room, it means that every mathematician was asleep in $1$ of $5$ private rooms at sometime. By pigeonhole principle, at least $6$ mathematicians were sleeping in the same private room, as $5\\times 5=25<26$ and no two of these $6$ mathematicians could have been asleep at the same time in that private room. <\/p>\n\n\n\n

Combining the above two cases completes the proof.<\/p>\n\n\n\n<\/div>\n","protected":false},"excerpt":{"rendered":"

During a lecture, each of  mathematicians falls asleep exactly once, and stays asleep for a nonzero amount of time. Each mathematician is awake at the moment the lecture starts, and the moment the lecture finishes. Prove that there are either  mathematicians such … Continue reading →<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"inline_featured_image":false},"categories":[13,10],"tags":[],"_links":{"self":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/5307"}],"collection":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=5307"}],"version-history":[{"count":7,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/5307\/revisions"}],"predecessor-version":[{"id":5329,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/5307\/revisions\/5329"}],"wp:attachment":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=5307"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=5307"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=5307"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}