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# [Core] GM

## GM

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 風之男 學院師父 發短消息 加為好友 當前離線 2# 大 中 小 發表於 2021-7-20 04:23 PM (第 138 天) 只看該作者 [顯示] [隱藏] Sidenote: What does GM mean? Presumably Geometry? 14, I'm not sure what the "correct" method is. But my first thought is that the diagram is not fixed, as in you can draw multiple different diagrams that are not congruent to each other, but still satisfy the condition (okay, that's not very convincing, isn't it...). What if you find a convenient triangle ABC? Say if ABC is a right-angled triangle with B as the right angle? Now it should be fairly trivial. I'll find the "correct" method later. Probably it's related to areas, ratios, base lengths, that kind of stuff. Related Geogebra to play around: https://www.geogebra.org/geometry/rkmaxhgt You are also putting faith in the question setter that the problem is actually solvable, which you should always. 39, This is supposed to be a sine law/cosine law problem, but you can choose not to go that route if you shoehorn the diagram onto a coordinate plane. Say D is the origin, DC and AD are the x- and y-axes. What would be the coordinates/equations for... well... everything? There is a problem quite similar in 2021DSE. The sine/cosine law route involves finding the area of triangle ADE two ways. AN being an altitude of triangle ADE is the big inspiration. Of course there is the problem of how long DE is, which you probably want to add some vertical/horizontal lines. Or... as I've attempted it, find DE using co-geom, and then find AN using areas... Whatever works, works - that's math. [ 本帖最後由 風之男 於 2021-7-20 04:49 PM 編輯 ] UID216418 帖子934 精華0 積分115 閱讀權限40 在線時間1639 小時 註冊時間2011-5-4 最後登錄2021-11-4  查看詳細資料 TOP

## 回覆 2# 風之男 的帖子

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 風之男 學院師父 發短消息 加為好友 當前離線 4# 大 中 小 發表於 2021-7-21 04:52 PM (第 137 天) 只看該作者 "Correct Method" for 14: As it turns out, this question is nothing more than just Pythagorean Theorem. You want to set quite a few unknowns for convenience. Let AD=a, ED=b, BD=c, DC=d. By the Pythagorean Theorem, you now have 3 equations in 4 unknowns. But don't worry, what you ultimately want to find is b^2+c^2 (and then sqrt it, but this is your "easy win"). If you look at your 3 equations carefully, you can add/subtract the equations to get to the answer of b^2+c^2=25. This also confirms my theory that the diagram is "not fixed", since it is impossible to have unique solutions with 3 equations and 4 unknowns. [ 本帖最後由 風之男 於 2021-7-21 04:59 PM 編輯 ] UID216418 帖子934 精華0 積分115 閱讀權限40 在線時間1639 小時 註冊時間2011-5-4 最後登錄2021-11-4  查看詳細資料 TOP
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