close
標題:
Find the length of BC.
發問:
A rectangle HFMO has sides HO=7 and OM=12. A triangle ABC has H as its orthocentre, O as its circumcentre, M as the midpoint of BC, and F as the foot of the altitude from A. Find the length of BC. 更新: Pak Wai sir, You mean orthocentre, centriod and circumcentre should be collinear ? Why ? 更新 2: Therefore, X lies on the height AF. ??(agree) Similarly prove for other sides with their medians, we have X lies on three heights.??(?, 3?) i.e. X is the orthocenter H and O, G, H are collinear. ?? (why X is H?) Anyway, I have checked that OGH are collinear due to the Euler line. Thx. 更新 3: Pak Wai sir 的解法無誤，只是 DSE 冇教，所以答了也不可能拿滿分。 反而 管理員 及 少年時 的可得滿分的。 無論如何，多謝三位的解說。 新年進步！
最佳解答:
其他解答:
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?
Find the length of BC.
發問:
A rectangle HFMO has sides HO=7 and OM=12. A triangle ABC has H as its orthocentre, O as its circumcentre, M as the midpoint of BC, and F as the foot of the altitude from A. Find the length of BC. 更新: Pak Wai sir, You mean orthocentre, centriod and circumcentre should be collinear ? Why ? 更新 2: Therefore, X lies on the height AF. ??(agree) Similarly prove for other sides with their medians, we have X lies on three heights.??(?, 3?) i.e. X is the orthocenter H and O, G, H are collinear. ?? (why X is H?) Anyway, I have checked that OGH are collinear due to the Euler line. Thx. 更新 3: Pak Wai sir 的解法無誤，只是 DSE 冇教，所以答了也不可能拿滿分。 反而 管理員 及 少年時 的可得滿分的。 無論如何，多謝三位的解說。 新年進步！
最佳解答:
此文章來自奇摩知識+如有不便請留言告知
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其他解答:
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?
文章標籤
全站熱搜
留言列表
