※ 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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