标签：elm orm glbp aic esb was oahp miss win
1. 凸四边形 $ABCD$ 中, $S_{\triangle{ABD}} + S_{\triangle{ABC}} = S_{\triangle{BCD}}$, $M$, $N$ 分别在 $AC$, $CD$ 上, $AM : AC = CN : CD$, 且 $B$, $M$, $N$ 共线, 求证: $M$, $N$ 分别为 $AC$, $CD$ 之中点.
解答:
考虑列出面积方程求解.
令$${AM \over AC} = {CN\over CD} = r,\ \triangle{ABC} = 1,\ \triangle{ABD} = x.$$ 需证明 $r = 1/2$. 由题意可得 $$\triangle{BCD} = x+1,\ \triangle{ABM} = r,$$ $$\triangle{ACN} = r\cdot\triangle{ACD} = r\cdot\left(\triangle{ABD} + \triangle{BCD} - \triangle{ABC}\right) = r\cdot (x + x+1 - 1) = 2rx,$$ $$\triangle{CMN} = \triangle{BCN} - \triangle{BCM} = r\cdot\triangle{BCD} - (1-r)\cdot\triangle{ABC}$$ $$= r\cdot (x+1) - (1-r) = r(x+2) - 1,$$ $$\triangle{AMN} = \triangle{ACN} - \triangle{CMN} = 2rx - r(x+2) + 1 = r(x-2) + 1,$$ $$\Rightarrow {\triangle{AMN}\over \triangle{ANC}} = r\Rightarrow {r(x-2) + 1\over 2rx} = r$$ $$\Rightarrow 2r^2x = rx - 2r + 1\Rightarrow 2r^2x - (x-2)r - 1 = 0$$ $$\Rightarrow (2r - 1)(rx + 1) = 0\Rightarrow r_1 = {1\over2},\ r_2 = -{1\over x}.$$ 由此得证.
2. 已知 $G$ 为 $\triangle{ABC}$ 内任一点, $BG$, $CG$ 分别交 $AC$, $AB$ 于点 $E$, $F$. 求使不等式 $S_{\triangle{BGF}}\cdot S_{\triangle{CGE}} \le kS^2_{\triangle{ABC}}$ 恒成立的 $k$ 的最小值.
解答:
考虑分别求出 ${\triangle{BGF} \over \triangle{ABC}}$ 及 ${\triangle{CGE}\over\triangle{ABC}}$, 再利用不等式求解.
令 ${AF\over AB} = x$, ${AE\over AC} = y$, $0 < x, y < 1$. 由面积关系得 $${\triangle{BGF} \over \triangle{ABC}} = {\triangle{BGF} \over \triangle{ABE}} \cdot {\triangle{ABE} \over \triangle{ABC}} = {BF \cdot BG \over AB\cdot BE}\cdot {AE\over AC},$$ 其中 ${BG\over BE}$ 可由 Menelaus 定理 ($CGF$ 截$\triangle{ABE}$) 得出 $${AF\over FB}\cdot{BG\over GE}\cdot{EC\over CA} = 1\Rightarrow {BG\over GE} = {1-x \over x}\cdot{1\over 1-y} = {1-x \over x-xy}\Rightarrow {BG\over BE} = {1-x \over 1-xy}.$$ 因此 $${\triangle{BGF} \over \triangle{ABC}} = (1-x)\cdot{1-x\over 1-xy}\cdot y = {y(1-x)^2 \over 1-xy}.$$ 同理可得 $${\triangle{CGE}\over\triangle{ABC}} = {x(1-y)^2 \over 1-xy}.$$ 因此 $${y(1-x)^2 \over 1-xy} \cdot {x(1-y)^2 \over 1-xy} = {xy(1 - x - y + xy)^2 \over (1-xy)^2} \le {xy(1 - 2\sqrt{xy} + xy)^2 \over (1-xy)^2},$$ 当 $x = y$ 时不等式取等号. 令$t = \sqrt{xy}$, 可得 $${t^2(1-t)^4 \over (1 - t^2)^2} = \left({t(1-t)^2 \over 1-t^2}\right)^2 \le \left({t(1-t)\over 1+t}\right)^2 = \left({-(1+t)^2 + 3t +1\over 1+t}\right)^2$$ $$=\left[ -\left(1+t + {2\over 1+t}\right) + 3\right]^2 \le \left(3-2\sqrt2\right)^2 = 17 - 12\sqrt2,$$ 当且仅当 $x = y = \sqrt2 - 1$ 时取等号.
综上, $k$ 的最小值为 $17-12\sqrt2$.
3. 正方形 $ABCD$ 的边长为 $2\sqrt{15}$, $E$, $F$ 分别是 $AB$, $BC$ 的中点, $AF$ 分别交 $DE$, $DB$ 于点 $M$, $N$, 则 $\triangle{DMN}$ 的面积是多少?
解答:
考虑求出 ${MN\over AF}$ 的值, 暨由 $\triangle{MND}$ 与 $\triangle{AFD}$ 面积比得出答案.
设正方形边长为 $2x$ ($x = \sqrt{15}$), 由已知可得 $${AN\over NF} = {AD\over BF} =2 \Rightarrow AN = {2\over3}AF = {2\sqrt5x \over 3}.$$ $$AM = AE\cdot \cos{\angle{EAM}} = x\cdot {2\over\sqrt5} = {2\sqrt5 \over 5}x.$$ 由此可得 $$MN = AN - AM = {4\sqrt5 \over 15}x$$ $$\Rightarrow {\triangle{MND}\over \triangle{AFD}} = {MN \over AF} = {4\over 15}$$ $$\Rightarrow \triangle{MND} = {4\over15}\cdot 2x^2 = 8.$$
4. 在 $\triangle{ABC}$ 中, $\angle{ABC} = 60^\circ$, 点 $O$, $H$ 分别为 $\triangle{ABC}$ 的外心, 垂心. 点 $D$, $E$ 分别在边 $BC$, $AB$ 上, 使 $BD = BH$, $BE = BO$. 已知 $BO = 1$, 求 $\triangle{BDE}$ 之面积.
解答:
考虑求出 $BD$, $BE$ 进而求 $\triangle{BDE}$ 之面积.
连结 $AO$ 并延长交 $\bigodot{O}$ 于 $F$, 连结 $CH$, $BF$, $CF$, $OC$. $$\begin{cases}BH \perp AC\\ CF\perp AC\end{cases}\Rightarrow BH\parallel CF,$$ $$\begin{cases}CH \perp AB\\ BF\perp AB\end{cases}\Rightarrow BF\parallel CH,$$ 因此 $BFCH$ 是平行四边形. $$\begin{cases}\angle{OFC} = \angle{ABC} = 60^\circ\\ OF = OC \end{cases} \Rightarrow OF = FC = OC.$$ 由以上可得: $$BD = BH = CF = OC = OB = BE\Rightarrow BE = ED = BD.$$ 即 $\triangle{BDE}$ 是正三角形, 因此其面积为 $\displaystyle{\sqrt3\over4}$.
5. 在平行四边形 $ABCD$ 中, $P$ 为 $\triangle{BAD}$ 内一点. 若 $S_{\triangle{PAB}} = 2$, $S_{\triangle{PCB}} = 5$, 求 $S_{\triangle{PBD}}$ 的值.
解答:
考虑使用共边定理求面积.
延长 $BP$交$AD$ 于 $F$, 连结 $AC$ 交 $PF$ 于 $E$.
由已知可得 $${\triangle{PBD} \over \triangle{ABP}} = {DF\over FA} = {AD -AF \over AF} = {AD \over AF} - 1 = {BC\over AF} - 1 = {CE \over EA} - 1= {\triangle{PBC} \over \triangle{APB}} - 1 = {3\over2}$$ $$\Rightarrow \triangle{PBD} = \triangle{ABP}\cdot{3\over2} = 2\cdot {3\over2} = 3.$$
另解: $$\triangle{PBD} = \triangle{PBC} + \triangle{PCD} - \triangle{BCD} = \triangle{PBC} + {1\over2}S_{ABCD} - \triangle{PAB} - {1\over2}S_{ABCD} = 3.$$
6. $C$, $D$ 是 $\bigodot{O}$ 上的点, $AD$, $BC$ 为 $\bigodot{O}$ 的切线, $AB$ 交 $\bigodot{O}$ 于 $K$, $L$ 且 $DL = CL$, $AC$ 与 $BD$ 交于 $P$, $M$ 为 $CD$ 中点. 求证: $M$, $P$, $K$ 共线.
解答:
如图所示, 考虑证明 $${\triangle{AMP} \over \triangle{BMP}} = {AK \over BK}.$$ 由圆幂定理可得: $${AK \over BK} = {{AD^2 \over AL} \over {BC^2 \over BL}} = {AD^2 \over BC^2}\cdot{BL\over AL}.$$ 再考虑左边面积比: $${\triangle{AMP} \over \triangle{BMP}} = {\triangle{AMP} \over \triangle{AMC}} \cdot {\triangle{BMD} \over \triangle{BMP}}\cdot{\triangle{AMC} \over \triangle{BMD}}$$ $$= {AP \over AC} \cdot {BD\over BP}\cdot{AD\over BC}.$$ 比较两边比例式可知, 需要证明 $${AD \over BC} \cdot {BL \over AL} = {AP \over AC} \cdot {BD\over BP}.$$ 首先, $${AL \over BL} = {AL \over DL} \cdot {CL \over BL} = {\sin\angle{ADL} \over \sin\angle{DAL}} \cdot {\sin\angle{CBL}\over \sin\angle{BCL}} = {\sin\angle{CBL}\over\sin\angle{DAL}}.$$ 其次, $${AP \over BP} = {\sin\angle{ABP} \over \sin\angle{BAP}}.$$ 因此 $${AL \over BL} \cdot {AP \over BP} = {\sin\angle{CBL}\over\sin\angle{DAL}} \cdot {\sin\angle{ABP} \over \sin\angle{BAP}} = {\sin\angle{CBL} \over \sin\angle{BAP}} \cdot {\sin\angle{ABP} \over \sin\angle{DAL}} = {AC\over BC} \cdot {AD \over BD},$$ 整理即为 $${AD \over BC} \cdot {BL \over AL} = {AP \over AC} \cdot {BD\over BP}.$$
标签：elm orm glbp aic esb was oahp miss win
原文地址：http://www.cnblogs.com/zhaoyin/p/6006348.html
