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# [讨论] 关于三角形的几个问题

楼主| 发表于 2014-7-28 23:47:12 | 显示全部楼层
 /forum.php ... 12&fromuid=1455 中得到了：求三角形内点与各个顶点构成的三个三角形，且三个三角形有等内切圆的最终结果 $$bc(-c+a-b)(a^3+3a^2b+3a^2c-ab^2+6abc-ac^2-3b^3-9b^2c-9bc^2-3c^3)(a^4-2a^3b-2a^3c-7a^2b^2-6a^2bc-7a^2c^2+2ab^3+22ab^2c+22abc^2+2ac^3+6b^4+36b^3c+60b^2c^2+36bc^3+6c^4)+2(a+b+c)(-c+a-b)(2a^6b+2a^6c-4a^5b^2+26a^5bc-4a^5c^2-16a^4b^3-31a^4b^2c-31a^4bc^2-16a^4c^3+8a^3b^4-122a^3b^3c-204a^3b^2c^2-122a^3bc^3+8a^3c^4+26a^2b^5+58a^2b^4c+24a^2b^3c^2+24a^2b^2c^3+58a^2bc^4+26a^2c^5-4ab^6+112ab^5c+332ab^4c^2+272ab^3c^3+332ab^2c^4+112abc^5-4ac^6-12b^7-45b^6c-117b^5c^2-258b^4c^3-258b^3c^4-117b^2c^5-45bc^6-12c^7)x+(16a^8+80a^7b+80a^7c-40a^6b^2+241a^6bc-40a^6c^2-368a^5b^3-398a^5b^2c-398a^5bc^2-368a^5c^3-40a^4b^4-697a^4b^3c-710a^4b^2c^2-697a^4bc^3-40a^4c^4+496a^3b^5+908a^3b^4c+1748a^3b^3c^2+1748a^3b^2c^3+908a^3bc^4+496a^3c^5+136a^2b^6+895a^2b^5c+2796a^2b^4c^2+4538a^2b^3c^3+2796a^2b^2c^4+895a^2bc^5+136a^2c^6-208ab^7-462ab^6c-966ab^5c^2-2108ab^4c^3-2108ab^3c^4-966ab^2c^5-462abc^6-208ac^7-72b^8-567b^7c-2430b^6c^2-5913b^5c^3-7956b^4c^4-5913b^3c^5-2430b^2c^6-567bc^7-72c^8)x^2+(4(a+b+c))(16a^6+45a^5b+45a^5c-29a^4b^2-168a^4bc-29a^4c^2-202a^3b^3-494a^3b^2c-494a^3bc^2-202a^3c^3-6a^2b^4+332a^2b^3c-292a^2b^2c^2+332a^2bc^3-6a^2c^4+221ab^5+561ab^4c+1234ab^3c^2+1234ab^2c^3+561abc^4+221ac^5-45b^6-276b^5c-579b^4c^2-792b^3c^3-579b^2c^4-276bc^5-45c^6)x^3+(656a^6+1344a^5b+1344a^5c-368a^4b^2+1900a^4bc-368a^4c^2-1664a^3b^3-2528a^3b^2c-2528a^3bc^2-1664a^3c^3-720a^2b^4-3080a^2b^3c-800a^2b^2c^2-3080a^2bc^3-720a^2c^4+320ab^5+3072ab^4c+7552ab^3c^2+7552ab^2c^3+3072abc^4+320ac^5+432b^6+3132b^5c+8496b^4c^2+12168b^3c^3+8496b^2c^4+3132bc^5+432c^6)x^4-96(a+b+c)(4a^4-14a^3b-14a^3c-10a^2b^2+50a^2bc-10a^2c^2+28ab^3+41ab^2c+41abc^2+28ac^3-8b^4-37b^3c-18b^2c^2-37bc^3-8c^4)x^5+(576a^4-144a^2bc-6048ab^2c-6048abc^2-576b^4-7056b^3c-7200b^2c^2-7056bc^3-576c^4)x^6+1728(a+b+c)(ab+ac-b^2-c^2)x^7+5184bcx^8=0$$ $$ca(-a+b-c)(-3a^3-a^2b-9a^2c+3ab^2+6abc-9ac^2+b^3+3b^2c-bc^2-3c^3)(6a^4+2a^3b+36a^3c-7a^2b^2+22a^2bc+60a^2c^2-2ab^3-6ab^2c+22abc^2+36ac^3+b^4-2b^3c-7b^2c^2+2bc^3+6c^4)+2(a+b+c)(-a+b-c)(-12a^7-4a^6b-45a^6c+26a^5b^2+112a^5bc-117a^5c^2+8a^4b^3+58a^4b^2c+332a^4bc^2-258a^4c^3-16a^3b^4-122a^3b^3c+24a^3b^2c^2+272a^3bc^3-258a^3c^4-4a^2b^5-31a^2b^4c-204a^2b^3c^2+24a^2b^2c^3+332a^2bc^4-117a^2c^5+2ab^6+26ab^5c-31ab^4c^2-122ab^3c^3+58ab^2c^4+112abc^5-45ac^6+2b^6c-4b^5c^2-16b^4c^3+8b^3c^4+26b^2c^5-4bc^6-12c^7)y+(-72a^8-208a^7b-567a^7c+136a^6b^2-462a^6bc-2430a^6c^2+496a^5b^3+895a^5b^2c-966a^5bc^2-5913a^5c^3-40a^4b^4+908a^4b^3c+2796a^4b^2c^2-2108a^4bc^3-7956a^4c^4-368a^3b^5-697a^3b^4c+1748a^3b^3c^2+4538a^3b^2c^3-2108a^3bc^4-5913a^3c^5-40a^2b^6-398a^2b^5c-710a^2b^4c^2+1748a^2b^3c^3+2796a^2b^2c^4-966a^2bc^5-2430a^2c^6+80ab^7+241ab^6c-398ab^5c^2-697ab^4c^3+908ab^3c^4+895ab^2c^5-462abc^6-567ac^7+16b^8+80b^7c-40b^6c^2-368b^5c^3-40b^4c^4+496b^3c^5+136b^2c^6-208bc^7-72c^8)y^2+(4(a+b+c))(-45a^6+221a^5b-276a^5c-6a^4b^2+561a^4bc-579a^4c^2-202a^3b^3+332a^3b^2c+1234a^3bc^2-792a^3c^3-29a^2b^4-494a^2b^3c-292a^2b^2c^2+1234a^2bc^3-579a^2c^4+45ab^5-168ab^4c-494ab^3c^2+332ab^2c^3+561abc^4-276ac^5+16b^6+45b^5c-29b^4c^2-202b^3c^3-6b^2c^4+221bc^5-45c^6)y^3+(432a^6+320a^5b+3132a^5c-720a^4b^2+3072a^4bc+8496a^4c^2-1664a^3b^3-3080a^3b^2c+7552a^3bc^2+12168a^3c^3-368a^2b^4-2528a^2b^3c-800a^2b^2c^2+7552a^2bc^3+8496a^2c^4+1344ab^5+1900ab^4c-2528ab^3c^2-3080ab^2c^3+3072abc^4+3132ac^5+656b^6+1344b^5c-368b^4c^2-1664b^3c^3-720b^2c^4+320bc^5+432c^6)y^4-96(a+b+c)(-8a^4+28a^3b-37a^3c-10a^2b^2+41a^2bc-18a^2c^2-14ab^3+50ab^2c+41abc^2-37ac^3+4b^4-14b^3c-10b^2c^2+28bc^3-8c^4)y^5+(-576a^4-7056a^3c-6048a^2bc-7200a^2c^2-144ab^2c-6048abc^2-7056ac^3+576b^4-576c^4)y^6+1728(a+b+c)(-a^2+ab+bc-c^2)y^7+5184cay^8=0$$ $$ab(-b+c-a)(-3a^3-9a^2b-a^2c-9ab^2+6abc+3ac^2-3b^3-b^2c+3bc^2+c^3)(6a^4+36a^3b+2a^3c+60a^2b^2+22a^2bc-7a^2c^2+36ab^3+22ab^2c-6abc^2-2ac^3+6b^4+2b^3c-7b^2c^2-2bc^3+c^4)+2(a+b+c)(-b+c-a)(-12a^7-45a^6b-4a^6c-117a^5b^2+112a^5bc+26a^5c^2-258a^4b^3+332a^4b^2c+58a^4bc^2+8a^4c^3-258a^3b^4+272a^3b^3c+24a^3b^2c^2-122a^3bc^3-16a^3c^4-117a^2b^5+332a^2b^4c+24a^2b^3c^2-204a^2b^2c^3-31a^2bc^4-4a^2c^5-45ab^6+112ab^5c+58ab^4c^2-122ab^3c^3-31ab^2c^4+26abc^5+2ac^6-12b^7-4b^6c+26b^5c^2+8b^4c^3-16b^3c^4-4b^2c^5+2bc^6)z+(-72a^8-567a^7b-208a^7c-2430a^6b^2-462a^6bc+136a^6c^2-5913a^5b^3-966a^5b^2c+895a^5bc^2+496a^5c^3-7956a^4b^4-2108a^4b^3c+2796a^4b^2c^2+908a^4bc^3-40a^4c^4-5913a^3b^5-2108a^3b^4c+4538a^3b^3c^2+1748a^3b^2c^3-697a^3bc^4-368a^3c^5-2430a^2b^6-966a^2b^5c+2796a^2b^4c^2+1748a^2b^3c^3-710a^2b^2c^4-398a^2bc^5-40a^2c^6-567ab^7-462ab^6c+895ab^5c^2+908ab^4c^3-697ab^3c^4-398ab^2c^5+241abc^6+80ac^7-72b^8-208b^7c+136b^6c^2+496b^5c^3-40b^4c^4-368b^3c^5-40b^2c^6+80bc^7+16c^8)z^2+(4(a+b+c))(-45a^6-276a^5b+221a^5c-579a^4b^2+561a^4bc-6a^4c^2-792a^3b^3+1234a^3b^2c+332a^3bc^2-202a^3c^3-579a^2b^4+1234a^2b^3c-292a^2b^2c^2-494a^2bc^3-29a^2c^4-276ab^5+561ab^4c+332ab^3c^2-494ab^2c^3-168abc^4+45ac^5-45b^6+221b^5c-6b^4c^2-202b^3c^3-29b^2c^4+45bc^5+16c^6)z^3+(432a^6+3132a^5b+320a^5c+8496a^4b^2+3072a^4bc-720a^4c^2+12168a^3b^3+7552a^3b^2c-3080a^3bc^2-1664a^3c^3+8496a^2b^4+7552a^2b^3c-800a^2b^2c^2-2528a^2bc^3-368a^2c^4+3132ab^5+3072ab^4c-3080ab^3c^2-2528ab^2c^3+1900abc^4+1344ac^5+432b^6+320b^5c-720b^4c^2-1664b^3c^3-368b^2c^4+1344bc^5+656c^6)z^4-96(a+b+c)(-8a^4-37a^3b+28a^3c-18a^2b^2+41a^2bc-10a^2c^2-37ab^3+41ab^2c+50abc^2-14ac^3-8b^4+28b^3c-10b^2c^2-14bc^3+4c^4)z^5+(-576a^4-7056a^3b-7200a^2b^2-6048a^2bc-7056ab^3-6048ab^2c-144abc^2-576b^4+576c^4)z^6+1728(a+b+c)(-a^2+ac-b^2+bc)z^7+5184abz^8=0$$

 第1个小题，使3个三角形周长相等的点，有一种特殊情况有简明的几何解。 当三角形ABC为直角三角形时（A为直角顶点），P点在三角形外，并使得四边形ABPC构成矩形。

### 评分

 关于第1小题，可以探讨两种特殊情况： 1.1 P点位于三角形的外接圆时，三角形的边长约束？是否必须为直角三角形形 2.2 P点位于三角形的一条边上时，三角形的边长约束？P点可内可外，在一条边上为一种临界状况，或许值得一探。

楼主| 发表于 2014-7-29 22:22:56 | 显示全部楼层
 对于hujunhua 回复的41#： 由于当$$ABC$$构成直角三角形时，$$a=x,b=y,c=z$$,此时有$$a+y+z=b+x+z=c+x+y=L$$,即$$L=a+b+c$$ 代入5#结果： $$-(-c+a+b)^2(-c+a-b)^2(c+a-b)^2-2(-c+a+b)(-c+a-b)(c+a-b)(a^2-2ab-2ac+b^2-2bc+c^2)L+ (-5a^4+4a^3b+4a^3c+2a^2b^2-4a^2bc+2a^2c^2+4ab^3-4ab^2c-4abc^2+4ac^3-5b^4+4b^3c+2b^2c^2+4bc^3-5c^4)L^2＝0$$ 化简得到：$$-8(a^2+b^2-c^2)(a^2-b^2-c^2)(a^2-b^2+c^2)＝0$$ 即可以断定当三角形$$ABC$$为直角三角形时,$$P$$点在三角形外，并使得四边形$$ABCP$$构成矩形 对于42#问题： 1.$$P,A,B,C$$共圆的条件为$$(ax+by-cz)(ax-by+cz)(-ax+by+cz)=0$$ ,不妨设$$ax+by=cz$$,代入$$a+y+z=b+x+z=c+x+y=L$$，得到$$L=\frac{-(a^2-2ab+b^2-c^2)}{-c+a+b}$$ 并代入5#结果化简得到：$$-4(-c+a-b)^2(c+a-b)^2(a^2+b^2-c^2)^2＝0$$ 显然得到$$a^2+b^2=c^2$$,即必须为直角三角形

楼主| 发表于 2014-7-30 21:56:18 | 显示全部楼层
 其实对于5#的结果： $$-(-c+a+b)^2(-c+a-b)^2(c+a-b)^2-(2(-c+a+b))(-c+a-b)(c+a-b)(a^2-2ab-2ac+b^2-2bc+c^2)L+(-5a^4+4a^3b+4a^3c+2a^2b^2-4a^2bc+2a^2c^2+4ab^3-4ab^2c-4abc^2+4ac^3-5b^4+4b^3c+2b^2c^2+4bc^3-5c^4)L^2=0$$ 解得： $L＝\frac{(c+a-b)(-c+a-b)(-c+a+b)}{2\sqrt{-a^4+2a^2b^2+2a^2c^2-b^4+2b^2c^2-c^4}\pm(a^2-2ab-2ac+b^2-2bc+c^2)}$ 若设 $p=\frac{a+b+c}{2}$ $S=\frac{\sqrt{2a^2b^2+2a^2c^2+2b^2c^2-a^4-b^4-c^4}}{4}=\frac{\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}}{4}＝pr$ $s=\frac{\sqrt{2ab+2ac+2bc-a^2-b^2-c^2}}{4}$ 则有 $L=\frac{16S^2}{(\pm 16s^2+8S)2p}=\frac{r}{1\pm \frac{2s^2}{S}}$ 若$$a^2+b^2=c^2$$,则有 $r=\frac{S}{p}=\frac{a+b-c}{2},S=\frac{ab}{2}$ 上式可进一步化简(第二种情形，取一）得到： $L=\frac{r}{1-\frac{2s^2}{S}}=\frac{2ab}{a+b-c}=\frac{2ab+a^2+b^2-c^2}{a+b-c}＝\frac{2S}{r}=a+b+c$

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