Find the length of BC.

HO // FM (rectangle properties)

Joint AM that intersects HO at the centroid G.
so, AM = 3GM
hence AF = 3FH = 36 (ΔAOH ~ ΔAFM since equilangular)
AH = AF - HF = 36 - 12 = 24

Joint OA and OC where OA = OB = OC = radius R

OA^2 = OH^2 + HA^2
OA = (7^2 + 24^2)^(1/2) = 25

OC = AO = 25

CM^2 = OC^2 - OM^2
CM = (25^2 - 12^2)^(1/2) = 481^(1/2)
BC = 2CM = 2[481^(1/2)]

2015-01-18 03:14:40 補充：
It should be ΔAHG ~ ΔAFM since equilangular
∠GAH = ∠MAF (common ∠)
∠AGH = ∠AMF (corr. ∠, GH//FM)
∠AHG = ∠AFM (corr. ∠, GH//FM)

2015-01-18 12:18:50 補充：
In this question, let G be the centroid.
Extend OG until a certain point, saids X, such that OG = 2GX.
GX = 2GO (we set this)
AG = 2MG since G is the centroid
∠OGM = ∠XGA (vert. opp. ∠s)
i.e. ΔOGM ~ ΔXGA (2 sides in ratio, int. ∠s)
(to be cont.)

2015-01-18 12:24:58 補充：
(It should be OG = (1/2)GX, typing error.)
We have
∠OMG = ∠OAX => AX // OM
But OM ⊥ BC => AX ⊥ BC
Therefore, X lies on the height AF.
Similarly prove for other sides with their medians, we have X lies on three heights.
i.e. X is the orthocenter H and O, G, H are collinear.

Note that 2OG = GH

2015-01-18 12:27:14 補充：
In fact, the line passes through the centroid, the circumcenter and the orthocenter of a triangle is called its Euler line.

2015-01-18 12:32:53 補充：
Sorry for my typing errors.
The last few lines should be

∠OMG = ∠XAG => AX // OM
But OM ⊥ BC => AX ⊥ BC

(again, sorry for my typing errors =])

2015-01-18 12:46:03 補充：
The graph could be drawn like:

http://i1379.photobucket.com/albums/ah127/jackiekwan1988/672A547D540D_zpsd107d3ef.png

2015-01-18 17:13:43 補充：
題目給了HO是歐拉線……就是不知道用不用證明它 0 0

2015-01-18 19:20:20 補充：
題目冇明示，不過有外心同垂心，好自然就諗咗去Euler line果邊 XD

2015-01-19 11:13:50 補充：
Similarly prove for other sides with their medians, we have X lies on three heights.⋯⋯(?, 3?)
It is to prove that the three heights of the triangle pass through X.
There are 3 heights, one as AH with the vertex A and the foot H of the height AH.

2015-01-19 11:21:06 補充：
Repeating this process for the same point X with different vertices (B, C) and their feet of the altitude, and you could prove that all the heights pass through X. Hence X is the orthocenter, where we usually called H.

2015-01-19 11:24:09 補充：
AF, not AH = =

2015-01-19 11:30:30 補充：
我經常打錯字…… QQ
Co-geom簡單易懂，可是沒有想到這邊。

對了，你們是用WORD打的嗎？
(比如這張：http://postimg.org/image/3ztr39vs1/)

2015-01-19 12:10:48 補充：
我是沒有用過這些數學軟件……QQ

Co-geom呢類係唔錯
http://postimg.org/image/3ztr39vs1/

2015-01-19 11:44:17 補充：
Re：少年時
謝謝 =)
let B嘅y-coordinate係y的確會快好多 XD

Re：Pak Wai
我用慣哥個叫 MathType
感覺上覺得幾 OK

貓sir, 題目似乎是講 Geometry, 但解決方法不一定要用 Geometry,
例如用 Co-geom 之類.

2015-01-19 11:04:22 補充：
一點即明，管理員的DSE成績一定好好了。不過方法可改進：

Let O be the origin, so, H=(0,7), F=(12,7), M=(12,0)
Let A=(-x,7), B=(12,y), C＝(12,-y)

As OA=OB, so,
x²＋7²＝12²＋y²
==> x²＋49＝144＋y² ⋯⋯ (i)

And BH⊥AC, so,
(y－7)/12 * (7＋y)/(-x－12) = -1
==> y²－49＝12x＋144 ⋯⋯ (ii)

2015-01-19 11:05:25 補充：
(i)＋(ii), get,
x²＝12x＋288
==> (x－24)(x＋12)＝0
==> x＝24 or -12 (rej)

Sub into (i), get y = √481

∴ BC＝2x＝2√481

2015-01-19 11:07:35 補充：
如果懂得 Euler line, Pak Wai 的解答是最方便快捷的.

其實我不是太熟悉 geometry (幾何學) 的課題。

或者看看 雨後 前輩 是否有空。

貓貓會答這個嗎?我也不太會...

開估啦!!!!!!!!

左圖畫好些是做到的。

2015-01-17 08:32:21 補充：
但依家係想計 BC 有幾長，唔係想量出 BC 有幾長。答案有開方符號的。

2015-01-18 18:27:29 補充：
其實題目冇提到乜野歐拉線, 而且呢題數我喺DSE啲mock卷揾出來.

照我所知佢地(中六)應該都唔知什麼是Euler line, 無論如何, 你的答案是正確的.

睇吓有冇其他高手可以唔使用Euler line來做吧.

http://oi57.tinypic.com/1hebsl.jpg (比例有出入)

左圖H不是orthocentre,但HFMO是長方形
右圖H是orthocentre,但HFMO不是長方形
△ABC要在圓周上,而O不能在△ABC之外
B,C,M,O已經定位,只能透過A點來定位F&H
問題有否出錯?