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有冇數學達人!求教DSE

有冇數學達人!求教DSE

我買左本 Maths mock paper操, 有冇人識尼題??
   

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[隱藏]
This is one way to do it: https://imgur.com/a/EvgSx30

There might be other ways, I'll try to look for them.

Interesting tidbit: Length of CM (half of what we're required to calculate), (1+sqrt5)/2, is the golden ratio.

[ 本帖最後由 風之男 於 2021-1-29 10:22 PM 編輯 ]

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好熟面口,樓主尼本係咪朗文個本?我都唔係好識><

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引用:
原帖由 風之男 於 2021-1-29 07:56 PM 發表
This is one way to do it: https://imgur.com/a/EvgSx30

There might be other ways, I'll try to look for them.

Interesting tidbit: Length of CM (half of what we're required to calculate), (1+sqrt5) ...
厲害,學下野先!勁過marking scheme

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引用:
原帖由 風之男 於 2021-1-29 07:56 PM 發表
This is one way to do it: https://imgur.com/a/EvgSx30

There might be other ways, I'll try to look for them.

Interesting tidbit: Length of CM (half of what we're required to calculate), (1+sqrt5) ...
Thank you for sharing

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引用:
原帖由 見習天使 於 2021-2-1 12:24 PM 發表
好熟面口,樓主尼本係咪朗文個本?我都唔係好識>
有解題影片啊!有冇興趣一齊課金買黎睇??

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So there's another way, although it's kind of cheating.

You can redraw the figure (or even, use the figure given, because it can't not be in scale in this particular question), and you'd find out AB is roughly 3.5 times of MN. Hence B is the correct option.
Now this isn't the best practice when it comes to understanding math, but this technique is sufficient in reaching the correct answer for geometry MC questions in general. Also, math is a lot about intuition and inspiration, so this isn't all bad to start.

EDIT: Oops, 1+sqrt5 is definitely not over 4 lmao.

[ 本帖最後由 風之男 於 2021-2-3 10:45 PM 編輯 ]

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引用:
原帖由 風之男 於 2021-2-3 04:22 AM 發表
So there's another way, although it's kind of cheating.

You can redraw the figure (or even, use the figure given, because it can't not be in scale in this particular question), and you'd find out A ...
hahaaaa i got you, thank you!  maybe it's cheating, but it's fast, good enough for MC questions

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HI!我喺度分享鴽絎H做法:
首先,你要知道以下兩個前提(如果你aim 5-5**,一定要不假思索咁寫出嚟)
1、圖1顯示邊長為rH等邊三角形三點座標
2、圖2顯示外接等邊三角形H圓H圓心座標
之後,我哋就引入直角座標,以A為原點,B位於x-軸之上,設O為圓心,AB長為r
咁樣,我哋利用以上兩個前提可以輕易得到圖3,O座標為 (r/2,sqrt(3)r/6),N座標為(3r/4+1,sqrt(3)r/4) (MN平行於AB好易證,因為係MC就唔證啦)
因為ON=OA,我哋可以得到
(r/2)^2+(sqrt(3)r/6)^2=(r/4+1)^2+(sqrt(3)r/4-sqrt(3)r/6)^2
我就唔會直接解依條方程,會將選項逐個代入去,睇屪鉽啱,結果係r=1+sqrt(5)
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哀吾生之須臾,羨長江之無窮

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