关于向量的数学题（英文），会哪道都行、告诉我😑
\u5965\u6570\u9898\u7528\u82f1\u8bed\u600e\u4e48\u8bf4Questions/Problems for International Mathematical Olympiad
IMO=\u56fd\u9645\u5965\u6797\u5339\u514b\u6570\u5b66\u7ade\u8d5b
\u5965\u6570\u9898=\u5965\u6797\u5339\u514b\u6570\u5b66\u7ade\u8d5b\u9898\u76ee~
\u5e0c\u671b\u56de\u7b54\u5bf9\u4f60\u6709\u5e2e\u52a9
Two numbers are in the tatio 2:3. When 4 is added to each number the ratio changes to 5:7.
The sum of the two original numbers is
\u3010\u4e24\u6570\u4e4b\u6bd4\u4e3a2\uff1a3\uff0c\u5f53\u6bcf\u4e2a\u6570\u52a0\u4e0a4\u65f6\uff0c\u65b0\u7684\u4e24\u6570\u4e4b\u6bd4\u4e3a5\uff1a7\uff0c\u6c42\u539f\u6765\u4e24\u6570\u4e4b\u548c\u3011
\u5047\u8bbe\u539f\u6570\u4e3ax y
x/y=2/3 >> 3x=2y >> x=2y/3
(x+4)/(y+4)=5/7 >>7x+28=5y+20 >>14y/3+28=15y/3+20
y/3=8
y=24
x=16
\u6240\u4ee5\u539f\u6765\u4e24\u6570\u548c\u4e3a24+16=40
\u5e0c\u671b\u56de\u7b54\u5bf9\u4f60\u6709\u5e2e\u52a9
合力=100(lb)
合力与80lb的夹角的正切 = 60/80 = 3/4,
合力与80lb 的夹角 = arctan(3/4).
2, 合力的平方 = (9)^2 + (14)^2 = 81 + 196 = 277,
合力=(277)^(1/2)[牛]
9牛在S上的分量 = 9*cos(60度）= 9/2，
9牛在E上的分量 = 9*sin(60度）= 9*3^(1/2)/2。
14牛在S上的分量= 14*cos(30度）= 7*3^(1/2)，
14牛在E上的分量=-14*sin(30度）= -7.
合力在S上的分量= 9/2 + 7*3^(1/2),
合力在E上的分量= 9*3^(1/2)/2 - 7
合力bearing is S 角度 W,
角度的正切 = [7 - 9*3^(1/2)/2]/[9/2 + 7*3^(1/2)] = [14 - 9*3^(1/2)]/[14*3^(1/2) + 9]
= [277*3^(1/2) - 9*14*4]/[196*3 - 81]
= [277*3^(1/2) - 504]/507.
角度 = arctan{[277*3^(1/2) - 504]/507}
the direction of the resultant force is bearing S arctan{[277*3^(1/2) - 504]/507} degree W.
绛旓細By AC=tAB, we have c-a=t(b-a).That is c=t(b-a)+a=tb+(1-t)a.
绛旓細1,鍚堝姏鐨勫钩鏂=锛60)^2 + (80)^2 = 3600 + 6400 = 10000,鍚堝姏=100(lb)鍚堝姏涓80lb鐨勫す瑙掔殑姝ｅ垏 = 60/80 = 3/4,鍚堝姏涓80lb 鐨勫す瑙 = arctan(3/4).2, 鍚堝姏鐨勫钩鏂 = (9)^2 + (14)^2 = 81 + 196 = 277,鍚堝姏=(277)^(1/2)[鐗沒9鐗涘湪S涓婄殑鍒嗛噺 = 9*cos(60搴︼級= 9...
绛旓細鏍规嵁鐭㈤噺鐩稿噺鐨鍚戦噺鍥撅紝yyyyye2鐨勭瓟妗堟槸姝ｇ‘鐨勩
绛旓細(2)鍚戦噺ED=鍚戦噺EO+鍚戦噺OD=鍚戦噺d-1/4鍚戦噺a-3/4鍚戦噺b 鍚戦噺EF=1/5鍚戦噺d-1/4鍚戦噺a 涓ょ嚎涓嶅瀭鐩
绛旓細鏁板瑕佹敞閲嶇嫭绔嬫濊冿紝鏈変簺棰樺鎯虫兂鎬昏兘鎯冲嚭鏉鐨勶紝鏁板鐨勯瓍鍔涗篃灏卞湪姝ゃ傛彁绀猴細锛1锛夊悜閲a路鍚戦噺b=2sin銆攛锛嬒/6锛-2cosx 鍖栫畝寰 f銆攛銆=2sin銆攛-蟺/6锛夛紙2锛夎2sin銆攛-蟺/6锛=t=6/5 cos銆2x锛嵪/3銆=1-2t^2 (浜屽嶈鍏紡)
绛旓細=鍚戦噺BC路(鍚戦噺CA+鍚戦噺AB)+鍚戦噺CA路鍚戦噺AB =鍚戦噺BC路鍚戦噺CB+鍚戦噺CA路鍚戦噺AB =-25+鍚戦噺CA路鍚戦噺AB =-25+6*4*锛-cosA锛変綑寮﹀畾鐞 cosA=(16+36-25)/(2*4*6)=9/16 -25+6*4*锛-cosA锛=-25+24*(-9/16)=-77/2 濡傛灉鎮ㄨ鍙垜鐨勫洖绛旓紝璇风偣鍑烩滈噰绾充负婊℃剰绛旀鈥,绁濆涔犺繘姝ワ紒
绛旓細a锛漼锛峹锛宐锛2x锛峺 x=a+b,y=2a+b 鈹倄鈹=鈭2 鈹倅鈹=鈭5 xy鐨勫す瑙捨=arccos(x路y/[鈹倄鈹傗攤y鈹俔)=arccos(3/鈭10)
绛旓細鍚戦噺AD+鍚戦噺BE+鍚戦噺CF =鍚戦噺AO+鍚戦噺OD+鍚戦噺BO+鍚戦噺OE+鍚戦噺CO+鍚戦噺OF =(鍚戦噺AO+鍚戦噺OF)+(鍚戦噺BO+鍚戦噺OD)+(鍚戦噺CO+鍚戦噺OE)=鍚戦噺AF+鍚戦噺BD+鍚戦噺CE =1/2(鍚戦噺AB+鍚戦噺BC+鍚戦噺CA)=0 2 鍚戦噺PA+鍚戦噺PB+鍚戦噺PC+鍚戦噺PD =鍚戦噺PO+鍚戦噺OA+鍚戦噺PO+鍚戦噺OB+鍚戦噺PO+鍚戦噺OC+鍚戦噺PO+鍚戦噺OD =4鍚戦噺PO...
绛旓細鏁板涓,鏃㈡湁澶у皬鍙堟湁鏂瑰悜鐨勯噺鍙仛鍚戦噺(涓庣煝閲忎笉鍚,娌℃湁璧风偣缁堢偣)(鑻辨枃:vector) 娉:鍦ㄧ嚎鎬т唬鏁颁腑鐨勫悜閲忔槸鎸噉涓疄鏁扮粍鎴愮殑鏈夊簭鏁扮粍,绉颁负n缁村悜閲忋偽=(a1,a2,鈥,an) 绉颁负n缁村悜閲.鍏朵腑ai绉颁负鍚戦噺伪鐨勭i涓垎閲忋 ("a1"鐨"1"涓篴鐨勪笅鏍,"ai"鐨"i"涓篴鐨勪笅鏍,鍏朵粬绫绘帹)銆 鍦–++涓,涔熸湁鍚戦噺銆 鍚戦噺(鎴...
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