Câu hỏi của bad end night

Question

cho tam giac ABC vuông tại C,BF và AE là 2 trung tuyến,r là bán kính đường tròn nội tiếp với AE=m,BF=n.Chứng minh rằng $\frac{r^2}{m^2+n^2}< \frac{1}{20}$

tìm giá trị lớn nhất của $\frac{r^2}{m^2+n^2}$

Hoàng Lê Bảo Ngọc · Accepted Answer

A B C F E Đặt AB = c , AC = b , BC = a . Ta có : $CE^2=AC^2+AE^2;BF^2=AB^2+AF^2$$\Rightarrow BF^2+CE^2=AB^2+AC^2+\left(AE^2+AF^2\right)=AB^2+AC^2+EF^2$$=AB^2+AC^2+\left(\frac{1}{2}BC\right)^2=BC^2+\frac{BC^2}{4}=\frac{5}{4}BC^2$$\Rightarrow m^2+n^2=\frac{5a^2}{4}$Lại có : $S_{\Delta ABC}=p.r=\frac{a+b+c}{2}.r\Rightarrow\frac{ab}{2}=\frac{\left(a+b+c\right).r}{2}$ $\Rightarrow r=\frac{ab}{a+b+c}$Mặt khác ta lại có : $ab\le\frac{a^2+b^2}{2}$ , $\left(a+b\right)^2=a^2+b^2+2ab>a^2+b^2$$\Rightarrow a+b>\sqrt{a^2+b^2}$ $\Rightarrow a+b+\sqrt{a^2+b^2}>2\sqrt{a^2+b^2}$ $\Rightarrow\frac{1}{a+b+\sqrt{a^2+b^2}}< \frac{1}{2\sqrt{a^2+b^2}}$Ta có : $r=\frac{ab}{a+b+c}=\frac{ab}{a+b+\sqrt{a^2+b^2}}< \frac{a^2+b^2}{4\sqrt{a^2+b^2}}=\frac{\sqrt{a^2+b^2}}{4}=\frac{c}{4}$ $\Rightarrow r^2< \frac{c^2}{16}\Rightarrow\frac{r^2}{m^2+n^2}< \frac{c^2}{16}.\frac{4}{5c^2}=\frac{1}{20}$ . Vậy $\frac{r^2}{m^2+n^2}< \frac{1}{20}$Ta có : $c^2=a^2+b^2\ge2ab$ $\Rightarrow\frac{r^2}{c^2}=\frac{a^2b^2}{\left(a+b+\sqrt{a^2+b^2}\right).c^2}\le\frac{a^2b^2}{2ab\left(a+b+\sqrt{a^2+b^2}\right)}=\frac{ab}{2\left(a+b+\sqrt{a^2+b^2}\right)}$$\le\frac{ab}{4ab\left(1+\sqrt{2}\right)^2}=\frac{1}{4\left(1+\sqrt{2}\right)^2}$ $\Rightarrow\frac{4r^2}{5c^2}\le\frac{1}{5\left(1+\sqrt{2}\right)^2}=\frac{3-2\sqrt{2}}{5}$$\Rightarrow\frac{r^2}{m^2+n^2}\le\frac{3-2\sqrt{2}}{5}$. Vậy Max $\left(\frac{r^2}{m^2+n^2}\right)=\frac{3-2\sqrt{2}}{5}\Leftrightarrow a=b\Leftrightarrow$tam giác ABC vuông cân tại A.Câu trả lời: A B C F E Đặt AB = c , AC = b , BC = a . Ta có : $CE^2=AC^2+AE^2;BF^2=AB^2+AF^2$$\Rightarrow BF^2+CE^2=AB^2+AC^2+\left(AE^2+AF^2\right)=AB^2+AC^2+EF^2$$=AB^2+AC^2+\left(\frac{1}{2}BC\right)^2=BC^2+\frac{BC^2}{4}=\frac{5}{4}BC^2$$\Rightarrow m^2+n^2=\frac{5a^2}{4}$Lại có : $S_{\Delta ABC}=p.r=\frac{a+b+c}{2}.r\Rightarrow\frac{ab}{2}=\frac{\left(a+b+c\right).r}{2}$ $\Rightarrow r=\frac{ab}{a+b+c}$Mặt khác ta lại có : $ab\le\frac{a^2+b^2}{2}$ , $\left(a+b\right)^2=a^2+b^2+2ab>a^2+b^2$$\Rightarrow a+b>\sqrt{a^2+b^2}$ $\Rightarrow a+b+\sqrt{a^2+b^2}>2\sqrt{a^2+b^2}$ $\Rightarrow\frac{1}{a+b+\sqrt{a^2+b^2}}< \frac{1}{2\sqrt{a^2+b^2}}$Ta có : $r=\frac{ab}{a+b+c}=\frac{ab}{a+b+\sqrt{a^2+b^2}}< \frac{a^2+b^2}{4\sqrt{a^2+b^2}}=\frac{\sqrt{a^2+b^2}}{4}=\frac{c}{4}$ $\Rightarrow r^2< \frac{c^2}{16}\Rightarrow\frac{r^2}{m^2+n^2}< \frac{c^2}{16}.\frac{4}{5c^2}=\frac{1}{20}$ . Vậy $\frac{r^2}{m^2+n^2}< \frac{1}{20}$Ta có : $c^2=a^2+b^2\ge2ab$ $\Rightarrow\frac{r^2}{c^2}=\frac{a^2b^2}{\left(a+b+\sqrt{a^2+b^2}\right).c^2}\le\frac{a^2b^2}{2ab\left(a+b+\sqrt{a^2+b^2}\right)}=\frac{ab}{2\left(a+b+\sqrt{a^2+b^2}\right)}$$\le\frac{ab}{4ab\left(1+\sqrt{2}\right)^2}=\frac{1}{4\left(1+\sqrt{2}\right)^2}$ $\Rightarrow\frac{4r^2}{5c^2}\le\frac{1}{5\left(1+\sqrt{2}\right)^2}=\frac{3-2\sqrt{2}}{5}$$\Rightarrow\frac{r^2}{m^2+n^2}\le\frac{3-2\sqrt{2}}{5}$. Vậy Max $\left(\frac{r^2}{m^2+n^2}\right)=\frac{3-2\sqrt{2}}{5}\Leftrightarrow a=b\Leftrightarrow$tam giác ABC vuông cân tại A.