可以作一个三角形，它的边平行并且等于已知三角形的中线

isee

求证：可以作一个三角形，它的边平行并且等于已知三角形的中线。

如图，即证$AE,BF,CD$可构成三角形，追问，若$\triangle ABC$的面积为1，则此三角形的面积为多少？

kuing

构成的三角形的中线跟原三角形的三边也有关系。
这个问题的结论就是三角形几何不等式里的“中线对偶定理”的依据了……

kuing

翻了一下，见 aoshoo.com/bbs1/dispbbs.asp?boardid=48&id … mp;page=0&star=2 的13楼

地狱的死灵

面积也用向量积玩一下（以下运算都是向量）：
4AE×BF
=(2AE)×(2BF)
=(AB+AC)×(BC+BA)
=AB×BC+AC×BC+AB×BA+AC×BA
=3AB×BC
即中线三角形面积是原三角形面积的3/4。

isee

把图直接帖来：

楼上理应加绝对值就，完美了，若直接用LaTeX代码就更美的，向量即"$\$\vv {AB}\$$"(打引号部分，发帖后即$\vv {AB}$）。




同样的，而，三中线构成三角形，用向量看，是显然的：

$2(\vv{AE}+\vv {BF}+ \vv {CD})=\vv {AB}+\vv{AC}+\vv{BA}+\vv{BC}+\vv{CA}+\vv{CB}=\vv 0$

hbghlyj

isee 发表于 2013-8-23 07:32
把图直接帖来：

图加载不出来http://www.aoshoo.com/bbs1/UploadFile/2010-1/20101311143685894.jpg

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hbghlyj

补一个图

hbghlyj

kuing 发表于 2013-8-22 14:42
构成的三角形的中线跟原三角形的三边也有关系。
这个问题的结论就是三角形几何不等式里的“中线对偶定理” ...

在这份资料第629页的题Triangle inequalities: medians and sides用到了“中线对偶定理”

SIAM Rev., 21(1979) 559.
Problem 79-19, A Triangle Inequality, by M. S. KLAmKIN (University of Alberta).
If $a_1, a_2, a_3$ and $m_1, m_2, m_3$ denote the sides and corresponding medians of a triangle, respectively, prove that
$$
\left(a_1^2+a_2^2+a_3^2\right)\left(a_1 m_1+a_2 m_2+a_3 m_3\right) \geq 4 m_1 m_2 m_3\left(a_1+a_2+a_3\right)
$$
SIAM Rev., 22(1980) 509-511.
Solution by the proposer.
To prove (1) as well as to give a dual inequality, we will use the known duality theorem that
$$
F\left(a_1, a_2, a_3, m_1, m_2, m_3\right) \geq 0 \Leftrightarrow F\left(m_1, m_2, m_3, \frac{3 a_1}{4}, \frac{3 a_2}{4}, \frac{3 a_3}{4}\right) \geq 0
$$
This follows immediately from the fact that the three medians $m_1, m_2, m_3$ of any triangle are themselves sides of a triangle with respective medians $3 a_1 / 4,3 a_2 / 4,3 a_3 / 4$. For a more general duality result, see [1].
We will now prove successively that
$$
\sum a_1^2 \sum a_1 m_1 \geq \sum a_1 \sum a_1^2 m_1 \geq 4 m_1 m_2 m_3 \sum a_1
$$
where the summations are to be understood as cyclic sums over the indices $1,2,3$. The right-hand inequality of (3) follows immediately from the known inequality
$$
a_1^2 m_1+a_2^2 m_2+a_3^2 m_3 \geq 4 m_1 m_2 m_3
$$
which was obtained by Bager [2] by first establishing its dual, i.e.,
$$
4\left\{a_1 m_1^2+a_2 m_2^2+a_3 m_3^2\right\} \geq 9 a_1 a_2 a_3
$$
However, if we make the substitutions
$$
4 m_1^2=2 a_2^2+2 a_3^2-a_1^2, \text { etc. }
$$
we obtain
$$
2 \sum a_1 \sum a_1^2 \geq 3\left\{3 a_1 a_2 a_3+\sum a_1^3\right\}
$$
which was obtained by Colins in 1870 [3, p.13].

hbghlyj

kuing 发表于 2013-8-22 14:42
构成的三角形的中线跟原三角形的三边也有关系。
这个问题的结论就是三角形几何不等式里的“中线对偶定理” ...

在这份资料第4页也用到“中线对偶定理”
Via the median - duality transforming an arbitrary triangle $A_1 A_2 A_3$ into one formed by its medians ([3, pp.109 - 111]), inequality (13) becomes
$$\tag{14}
\left(\frac{h_1}{\sqrt{m_2 m_3}}\right)^t+\left(\frac{h_2}{\sqrt{m_3 m_1}}\right)^t+\left(\frac{h_3}{\sqrt{m_1 m_2}}\right)^t \leq 3 .
$$
Finally, in (12), we put $\lambda_i=\frac{1}{a_i^t}$ for $i=1,2,3$. A short calculation gives
$$\tag{15}
3\left(\frac{R}{F}\right)^{\frac{t}{2}} \geq \sum_{i=1}^3\left(\frac{\sqrt{a_i}}{m_i}\right)^t
$$
Here, we make use of the identity $a_1 a_2 a_3=4 R F$, where $F$ denotes the area of $A_1 A_2 A_3$
The median - dual of this inequality in turn reads
$$\tag{16}
\sum_{i=1}^3\left(\frac{\sqrt{m_i}}{a_i}\right)^t \leq 3\left(\frac{\sqrt{m_1 m_2 m_3}}{2 F}\right)^t
$$
Of course, if in (12) had we put $\lambda_i=\frac{\mu_i}{a_i^t}$ with $\mu_i>0, i=1,2,3$, we would obtain an even more general but less elegant inequality.

hbghlyj

kuing 发表于 2013-8-22 14:42
构成的三角形的中线跟原三角形的三边也有关系。
这个问题的结论就是三角形几何不等式里的“中线对偶定理” ...

在这份资料第218页也用到“中线对偶定理”
Mathematics Magazine , Sep., 1976, Vol. 49, No. 4 (Sep., 1976), pp. 211-218.pdf (606.6 KB, Downloads: 46)

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Q638. Let $a, b$, and $c$ denote the sides of an arbitrary triangle with respective medians $m_a, m_b$, and $m_c$. Determine all integral $p$ and $q$ so that
$$
\left(\frac{\sqrt{3}}{2}\right)^p\left(a^p m_a^q+b^p m_b^q+c^p m_c^q\right) \geqq\left(\frac{\sqrt{3}}{2}\right)^q\left(a^q m_a^p+b^q m_b^p+c^q m_c^p\right) .
$$
[Murray S. Klamkin, University of Waterloo.]
Q638. It is known that the medians $m_a, m_b, m_c$ form a triangle with respective medians $3 a / 4,3 b / 4$, $3 c / 4$. Consequently for any side-median inequality in the terms $a, b, c, m_a, m_b, m_c$ we have a dual median-side inequality in the terms $m_a, m_b, m_c, 3 a / 4,3 b / 4,3 c / 4$. Dualizing the inequality of the problem merely reverses the inequality sign, producing equality. Only $p=q$ yields this identity. (Note that $p$ and $q$ need not be integers.)

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产生的PDF(607kB)比原PDF(492kB)大, 但是看上去一样

hbghlyj

kuing 发表于 2013-8-22 14:42
构成的三角形的中线跟原三角形的三边也有关系。
这个问题的结论就是三角形几何不等式里的“中线对偶定理” ...

在Triangle inequalities from the triangle inequality, Klamkin, M.S. Elemente der Mathematik第54页也用到“中线对偶定理”
\begin{array}l\left(a_1^3 M_1, a_2^3 M_2, a_3^3 M_3\right) &\text{form a triangle.}
\\\left(a_1 M_1^3, a_2 M_2^3, a_3 M_3^3\right) &\text{form a triangle.}\end{array}The latter two inequalities are duals, i.e., if $\mathrm{F}\left(a_1, a_2, a_3, M_1, M_2, M_3\right) \geq 0$ is a valid inequality, so is $\mathrm{F}\left(M_1, M_2, M_3, 3 / 4 a_1, 3 / 4 a_2, 3 / 4 a_3\right) \geq 0$.

hbghlyj

isee 发表于 2013-8-23 07:32
$2(\vv{AE}+\vv {BF}+ \vv {CD})=\vv {AB}+\vv{AC}+\vv{BA}+\vv{BC}+\vv{CA}+\vv{CB}=\vv 0$

按这样也可以证明“若$D,E,F$为$BC,CA,AB$的$k:1$分点, 则$AD,BE,CF$构成三角形”吧
$\vv{AD}+\vv {BE}+ \vv {CF}\begin{aligned}[t]&=(1-k)\vv {AB}+k\vv{AC}\\&+(1-k)\vv {BC}+k\vv{BA}\\&+(1-k)\vv {CA}+k\vv{CB}=(1-2k)\vv {AB}+(1-2k)\vv{BC}+(1-2k)\vv{CA}=\vv 0\end{aligned}$