→ hahaj6u4503:得到垂直之後, 是圓的切線可以直接用圓冪證 EK = EL: 07/13 23:09
→ hahaj6u4503:EK^2 = EC * EC' = EL^2 07/13 23:10
※ 引述《FAlin (FA（バルシェ應援）)》之銘言：
: 4. Find all functions f: Z→Z, such that for all a+b+c = 0 holds :
: f(a)^2 + f(b)^2 + f(c)^2 = 2f(a)f(b) + 2f(b)f(c) + 2f(c)f(a)
: 5. Let △ABC be a triangle with ∠C = 90度 and that D be the foot of the
: altitude from C. Let X be a point in the interior of the segment CD.
: Let K be the point on the segment AX such that BK = BC. Similarly,
: let L be the point on the segment BX such that AL = AC.
: Let M be the point of intersection of AL and BK.
: Show that MK = ML.
: 6. Determine all positive integers n for which there exist non-negative
: integers a_1, a_2, ...,a_n such that:
: 1 1 1 1 2 n
: ----- + ----- + ... + ----- = ----- + ----- + ... + ----- = 1
: 2^a_1 2^a_2 2^a_n 3^a_1 3^a_2 3^a_n
Q5 防個雷
分別延長BX與AX，過A點做垂直BX的線與過B點垂直AX的線交於E點
因為AE⊥BX、BE⊥AX，所以X是△ABE的垂心
所以ECXD四點共線
連DL，由子母相似，可知AC^2 = AL^2 = AD * AB
所以△ALD ~ △ABL ∴∠ABL = ∠ALD
而∠AED = 90 - ∠EAD = 90 - ∠ABL = ∠ALD
∴AELD四點共圓，同理BEKD四點共圓
∴∠ALE = ∠ADE = 90 、 ∠EKB = ∠EDB = 90
最後
EK^2 = BE^2 - BK^2 = ED^2 + BD^2 - BK^2
= ED^2 + BD^2 - BC^2 = ED^2 - CD^2
EL^2 = AE^2 - AL^2 = ED^2 + AD^2 - AL^2
= ED^2 + AD^2 - AC^2 = ED^2 - CD^2
則 ME = ME 、 ∠EKM = ∠ ELM = 90度 、 KE = LE
∴△EKM ≡ △ELM
∴MK = ML 證畢
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◆ From: 140.112.244.138
※ 編輯: FAlin 來自: 140.112.244.138 (07/12 16:56)