题目

查看数据源：598a7f3740b385000915c47d

已知四边形 $ABCD$ 是正方形，$P$ 是边 $CD$ 上一点（点 $P$ 不与顶点重合），延长 $AP$ 与 $BC$ 的延长线交于点 $Q$．设 $\triangle ABQ,\triangle PAD,\triangle PCQ$ 的内切圆半径分别是 $r_{1},r_{2},r_{3}$．

【难度】

【出处】

2016年全国高中数学联赛湖南省预赛

【标注】

证明：$r_{1}^{2}\geqslant 4r_{2}r_{3}$，并指出点 $P$ 在什么位置时等号成立；
标注
答案

略

解析

如图所示，因为 $\triangle ABQ\sim \triangle PDA\sim \triangle PCQ$，所以\[r_{1}:r_{2}:r_{3}=AB:PD:PC.\]而\[AB=CD=PD+PC,\]故\[r_{1}=r_{2}+r_{3}\geqslant 2\sqrt{r_{2}r_{3}},\]即\[r_{1}^{2}\geqslant 4r_{2}r_{3}.\]当且仅当 $r_{2}=r_{3}$，即 $P$ 为 $CD$ 的中点时等号成立．
若 $AB=1$，试求证：$3-2\sqrt 2<r_{1}^{2}+r_{2}^{2}+r_{3}^{2}<\dfrac{1}{2}$．
标注
答案

略

解析

由第 $(1)$ 小题得 $r_{1}=r_{2}+r_{3}$，所以有\[\begin{split}r_{1}^{2}+r_{2}^{2}+r_{3}^{2}&=(r_{2}+r_{3})^{2}+r_{2}^{2}+r_{3}^{2}\\&=\left(r_{2}+r_{2}\cdot \dfrac{1-PD}{PD}\right)^{2}+r_{2}^{2}+r_{2}^{2}\left(\dfrac{1-PD}{PD}\right)^{2}\\&=r_{2}^{2}\left[\dfrac{1}{PD^{2}}+1+\left(\dfrac{1}{PD}-1\right)^{2}\right]\\&=r_{2}^{2}\cdot 2\cdot \left(\dfrac{1}{PD^{2}}-\dfrac{1}{PD}+1\right)\\&=\left(\dfrac{1-PD-AP}{2}\right)^{2}\times 2\left(\dfrac{1}{PD^{2}}-\dfrac{1}{PD}+1\right)\\&=\dfrac{(1+PD-\sqrt{1+PD^{2}})^{2}}{2}\cdot \left(\dfrac{1}{PD^{2}}-\dfrac{1}{PD}+1\right).\end{split}\]记 $\angle DAP =\theta ,0<\theta <\dfrac{\pi}{4}$，则\[\begin{split}r_{1}^{2}+r_{2}^{2}+r_{3}^{2}&=\dfrac{1}{2}\left(1+\tan\theta-\dfrac{1}{\cos\theta}\right)^{2}(1+\cot^{2}\theta -\cos \theta )\\&=\dfrac{1}{2}\cdot \dfrac{(\cos\theta +\sin\theta -1)^{2}}{\cos^{2}\theta }\cdot \dfrac{\sin^{2}\theta +\cos^{2}\theta -\sin\theta \cos\theta }{\sin^{2}\theta}\\&=\dfrac{1}{2}\cdot \dfrac{(\cos\theta +\sin\theta -1)^{2}(1-\sin\theta\cos\theta )}{(\sin\theta\cos\theta)^{2}}.\end{split}\]令 $\sin\theta +\cos\theta =t$，则 $1<t<\sqrt 2$，且 $\sin\theta \cos\theta =\dfrac{t^{2}-1}{2}$．故\[r_{1}^{2}+r_{2}^{2}+r_{3}^{2}=\dfrac{1}{2}\cdot \dfrac{(t-1)^{2}\left(1-\dfrac{t^{2}-1}{2}\right)}{\left(\dfrac{t^{2}-1}{2}\right)^{2}}=\dfrac{3-t^{2}}{(t+1)^{2}}.\]它是关于 $t$ 的单调递减函数．
所以，\[3-2\sqrt 2=\dfrac{3-(\sqrt 2)^{2}}{(\sqrt 2+1)^{2}}<r_{1}^{2}+r_{2}^{2}+r_{3}^{2}<\dfrac{3-1^{2}}{(1+1)^{2}}=\dfrac{1}{2}.\]即\[3-2\sqrt 2<r_{1}^{2}+r_{2}^{2}+r_{3}^{2}<\dfrac{1}{2}.\]

题目问题1 答案1 解析1 备注1 问题2 答案2 解析2

由第 $(1)$ 小题得 $r_{1}=r_{2}+r_{3}$，所以有\[\begin{split}r_{1}^{2}+r_{2}^{2}+r_{3}^{2}&=(r_{2}+r_{3})^{2}+r_{2}^{2}+r_{3}^{2}\\&=\left(r_{2}+r_{2}\cdot \dfrac{1-PD}{PD}\right)^{2}+r_{2}^{2}+r_{2}^{2}\left(\dfrac{1-PD}{PD}\right)^{2}\\&=r_{2}^{2}\left[\dfrac{1}{PD^{2}}+1+\left(\dfrac{1}{PD}-1\right)^{2}\right]\\&=r_{2}^{2}\cdot 2\cdot \left(\dfrac{1}{PD^{2}}-\dfrac{1}{PD}+1\right)\\&=\left(\dfrac{1-PD-AP}{2}\right)^{2}\times 2\left(\dfrac{1}{PD^{2}}-\dfrac{1}{PD}+1\right)\\&=\dfrac{(1+PD-\sqrt{1+PD^{2}})^{2}}{2}\cdot \left(\dfrac{1}{PD^{2}}-\dfrac{1}{PD}+1\right).\end{split}\]记 $\angle DAP =\theta ,0<\theta <\dfrac{\pi}{4}$，则\[\begin{split}r_{1}^{2}+r_{2}^{2}+r_{3}^{2}&=\dfrac{1}{2}\left(1+\tan\theta-\dfrac{1}{\cos\theta}\right)^{2}(1+\cot^{2}\theta -\cos \theta )\\&=\dfrac{1}{2}\cdot \dfrac{(\cos\theta +\sin\theta -1)^{2}}{\cos^{2}\theta }\cdot \dfrac{\sin^{2}\theta +\cos^{2}\theta -\sin\theta \cos\theta }{\sin^{2}\theta}\\&=\dfrac{1}{2}\cdot \dfrac{(\cos\theta +\sin\theta -1)^{2}(1-\sin\theta\cos\theta )}{(\sin\theta\cos\theta)^{2}}.\end{split}\]令 $\sin\theta +\cos\theta =t$，则 $1<t<\sqrt 2$，且 $\sin\theta \cos\theta =\dfrac{t^{2}-1}{2}$．故\[r_{1}^{2}+r_{2}^{2}+r_{3}^{2}=\dfrac{1}{2}\cdot \dfrac{(t-1)^{2}\left(1-\dfrac{t^{2}-1}{2}\right)}{\left(\dfrac{t^{2}-1}{2}\right)^{2}}=\dfrac{3-t^{2}}{(t+1)^{2}}.\]它是关于 $t$ 的单调递减函数．<br/>所以，\[3-2\sqrt 2=\dfrac{3-(\sqrt 2)^{2}}{(\sqrt 2+1)^{2}}<r_{1}^{2}+r_{2}^{2}+r_{3}^{2}<\dfrac{3-1^{2}}{(1+1)^{2}}=\dfrac{1}{2}.\]即\[3-2\sqrt 2<r_{1}^{2}+r_{2}^{2}+r_{3}^{2}<\dfrac{1}{2}.\]