内切圆距离平方和

Erstellt am: 10. Januar 2025

Beantwortet mit O1 von Chat01

Frage

在 $\Delta ABC$ 中， $\angle A$ 、 $\angle B$ 、 $\angle C$ 所对的边分别为 $a, b, c$ ，且 $c=10$ ， $\frac{\cos A}{\cos B} = \frac{b}{a} = \frac{4}{3}$ ， $P$ 为 $\Delta ABC$ 内切圆上的动点，求点 $P$ 到顶点 $A$ 、 $B$ 、 $C$ 的距离的平方和的最大值和最小值。

Gedanke

Setting up the problem

OK, let's work through triangle ABC with c = 10. Given $\frac{\cos A}{\cos B} = \frac{4}{3}$ , I’m mapping out $a$ and $b$ in terms of $k$ , then calculating $a$ , $b$ , and $c$ .

Breaking down constraints

OK, let's break down the constraints for $x$ . It needs to satisfy $10/7 < x < 10$ for side lengths $a$ , $b$ , and $c$ to form a valid triangle.

Verifying the ratio

I’m checking if $\frac{\cos A}{\cos B} = \frac{4}{3}$ aligns with $\frac{b}{a} = \frac{4}{3}$ . Using the law of cosines and side lengths, the equations match up.

Validating the equation

OK, let me see. I’m confirming that $\frac{\cos A}{\cos B} = \frac{4}{3}$ holds true by simplifying and rearranging the equation to ensure it aligns with the given ratio.

Verifying the setup

It's interesting to see that the triangle is indeed a right triangle, with $a^2 + b^2 = c^2$ , and $c$ as the hypotenuse. Given $c = 10$ , then $a^2 + b^2 = 100$ .

Drawing the triangle

I'm piecing together a 6-8-10 right triangle with the hypotenuse as 10. The angles align perfectly with these sides, and the incenter is 2 units away from each side.

Mapping coordinates

I'm placing the 6-8-10 triangle on the coordinate plane: $C = (0,0)$ , $B = (8,0)$ , and $A = (0,6)$ . The incircle has a radius of 2 and is centered at the incenter I.

Mapping the incenter

I'm gathering info on the incenter of a 6-8 right triangle. With coordinates $C = (0,0)$ , $B = (8,0)$ , and $A = (0,6)$ , the incenter is $(2,2)$ with a radius of 2.

Pinpointing incenter coordinates

Refining the incenter of a 6-8 triangle as $(2,2)$ , it's positioned 2 units from each side, matching the side lengths and ensuring the right angle at the origin aligns with the coordinate system.

Determining the incenter

OK, let me see. I’m realizing that in a right triangle, the incenter is calculated using $(a+b-c)/2$ . With $a = 6$ , $b = 8$ , and $c = 10$ , it’s at $(2,2)$ .

Summing distance squares

I’m thinking through $PA^2$ , $PB^2$ , and $PC^2$ by defining their coordinates. We’ll then find the minimum and maximum of their sum within the incircle's boundary.

Piecing it together

I’m combining terms to express $S = 3(x^2 + y^2) - 16x - 12y + 100$ . To find the minimum and maximum, I’ll consider points within and on the incircle.

Expressing S

Rewritten in terms of $(x-2)$ and $(y-2)$ , I express $S$ as $3((x-2)^2 + (y-2)^2) - 4x + 76$ . This adjustment simplifies the analysis.

Verifying calculations

Mapping out $S$ involves checking arithmetic. $S = 3((x-2)^2 + (y-2)^2) - 4x + 76$ , and ensuring terms are accurately combined, aligns with the incircle constraint.

Identifying critical points

Given $f(X,Y) = 3(X^2 + Y^2) - 4X$ , setting the partial derivatives to zero reveals critical points, like $(2/3, 0)$ , within the region $X^2 + Y^2 \leq 4$ .

Verifying critical point

I’m checking if $(2/3,0)$ is inside the circle $X^2 + Y^2 \leq 4$ . It’s a candidate for the minimum of $f(X,Y)$ and the corresponding $S$ .

Identifying boundary extremes

I'm noting that on the boundary $X^2 + Y^2 = 4$ , $f(X,Y)$ peaks when $X = -2$ , yielding $20$ and an $S$ of $88$ .

Identifying extrema

The global minimum of $S$ is $200/3$ at the interior point $(2/3,0)$ , while the global maximum is 88 at the boundary point $(2,0)$ . Progressing through these points confirms their alignment with the incircle boundary.

Confirming the boundary point

I verified that the maximum $S$ is 88 at the boundary point $(0,2)$ . This ensures the leftmost point on the incircle and aligns with our earlier findings.

Confirming the boundary

I verified the distance from $(0,2)$ to points $A$ , $B$ , and $C$ , ensuring it aligns with the incircle's boundary.

Verifying calculations

I checked the distances from $(8/3, 2)$ to $A$ , $B$ , and $C$ , confirming the sum of squares as $200/3$ , the global minimum.

Thought about 三角形内切圆，持续 1m 7s

Antwort

思路概要：

确定三角形边长与形状
已知 $c=10$ 且
$\frac{\cos A}{\cos B} \;=\; \frac{b}{a} \;=\; \frac{4}{3},$
又由对边记号可知 $a=BC,\;b=AC,\;c=AB$ 。将这一比值条件结合到余弦定理中，可以推导出
$a^2 + b^2 \;=\; c^2.$
因而 $\triangle ABC$ 必是直角三角形，且直角在 $C$ 。给定 $c=10$ ，又有
$\frac{b}{a} \;=\; \frac{4}{3},\quad a^2 + b^2 \;=\; 10^2=100.$
令 $a=3k,\, b=4k$ ，则
$3^2 k^2 \;+\; 4^2 k^2 \;=\; 9k^2 + 16k^2 \;=\; 25k^2 \;=\; 100,$
可得 $k=2$ ，从而
$a=6,\quad b=8,\quad c=10.$
故三角形的三边为 $6,8,10$ ，并且 $\angle C=90^\circ$ 。
建立坐标系并确定顶点与内心
为方便计算，常将直角顶点 $C$ 放在原点 $(0,0)$ ，令
$B=(8,\,0),\quad A=(0,\,6),\quad C=(0,\,0).$
对于直角三角形，内心（即内切圆圆心）在“直角顶点”与两条直角边上的距离都等于内切圆半径 $r$ 。
- 该三角形的半周长为 $s \;=\; \frac{a+b+c}{2} \;=\; \frac{6+8+10}{2} \;=\; 12,$ 内切圆半径 $r \;=\; \frac{\text{三角形面积}}{s} \;=\; \frac{\tfrac12\times 6\times 8}{12} \;=\; 2.$
- 另外，对“直角在原点、两条直角边沿坐标轴”的情形，可知内心坐标是 $(r,r)$ ，即 $(2,2)$ 。
  因而内切圆方程为
$(x-2)^2 \;+\; (y-2)^2 \;=\; 2^2 \;=\; 4.$
设点 $P=(x,y)$ 在内切圆上，表达目标函数
我们要在该内切圆上，求
$PA^2 \;+\; PB^2 \;+\; PC^2$
的最大值与最小值。
- 三个顶点坐标： $A=(0,6),\quad B=(8,0),\quad C=(0,0).$
- 因而 $PA^2 \;=\; (x-0)^2 + (y-6)^2 \;=\; x^2 + (y-6)^2,$ $PB^2 \;=\; (x-8)^2 + (y-0)^2 \;=\; (x-8)^2 + y^2,$ $PC^2 \;=\; (x-0)^2 + (y-0)^2 \;=\; x^2 + y^2.$
将它们相加：
$S \;=\; PA^2 + PB^2 + PC^2 \;=\; \bigl[x^2 + (y-6)^2\bigr] + \bigl[(x-8)^2 + y^2\bigr] + \bigl[x^2 + y^2\bigr].$
展开并合并同类项可得
$S \;=\; 3x^2 \;+\; 3y^2 \;-\;16x \;-\;12y \;+\; 100.$
转化为以圆心为坐标原点的形式，便于在圆上寻找极值
令内心 $I=(2,2)$ ，则令
$X \;=\; x-2,\quad Y \;=\; y-2,$
内切圆即 $X^2 + Y^2 = 4$ 。把 $x=X+2$ 、 $y=Y+2$ 代入上式，可化简得到
$S \;=\; 3\bigl[(X+2)^2 + (Y+2)^2\bigr] \;-\;16\,(X+2) \;-\;12\,(Y+2) \;+\; 100 \;\;=\;\; 3(X^2 + Y^2) \;-\; 4X \;+\; 68.$
（详细推导略，关键在于将 $x^2+y^2$ 改写成 $(X+2)^2 + (Y+2)^2$ ，然后合并。）

因此我们得到一个在圆 $X^2+Y^2 \le 4$ 上的“二元二次函数”
$f(X,Y) \;=\; 3(X^2 + Y^2) \;-\; 4X, \quad S = f(X,Y) + 68.$
求极值：先看圆内部临界点，再看圆周
- （1）圆内部的临界点
  对
  $f(X,Y) = 3(X^2 + Y^2) - 4X$
  做偏导数并设为 0：
  $\frac{\partial f}{\partial X} = 6X - 4 = 0 \;\;\Longrightarrow\;\; X = \tfrac{2}{3}, \quad \frac{\partial f}{\partial Y} = 6Y = 0 \;\;\Longrightarrow\;\; Y = 0.$
  故唯一的驻点是 $\bigl(\tfrac{2}{3},\,0\bigr)$ 。它是否落在圆内？
  $\left(\frac{2}{3}\right)^2 + 0^2 = \frac{4}{9} \;<\;4,$
  确实在圆的内部。将其带回 $f(X,Y)$ ：
  $f\!\Bigl(\tfrac{2}{3},\,0\Bigr) = 3\left(\tfrac{4}{9}\right) \;-\;4\!\left(\tfrac{2}{3}\right) = \tfrac{4}{3} \;-\; \tfrac{8}{3} = -\tfrac{4}{3},$
  因而
  $S_{\text{内点}} = f\!\Bigl(\tfrac{2}{3},\,0\Bigr) + 68 = -\tfrac{4}{3} + 68 = 68 - \tfrac{4}{3} = \tfrac{200}{3}.$
  由于 $f(X,Y)$ 中二次项系数为正， $\bigl(\tfrac{2}{3},\,0\bigr)$ 应该给出全局最小值。
- （2）在圆周 $X^2+Y^2=4$ 上的取值
  当 $X^2+Y^2=4$ 时，
  $f(X,Y) = 3\cdot 4 \;-\;4X = 12 - 4X.$
  要想在周线上最大化 $f$ ，需让 $-4X$ 尽量大，即 $X$ 越小越好；由于 $X^2\le 4$ ，最小可到 $X=-2$ 。这时 $Y=0$ ，故
  $f(-2,0) = 12 - 4(-2) = 12 + 8 = 20, \quad S_{\max} = 20 + 68 = 88.$
  而要在周线上最小化 $f$ ，需让 $-4X$ 尽量小，即 $X$ 最大到 $+2$ 。此时
  $f(2,0) = 12 - 4(2) = 4, \quad S = 4 + 68 = 72.$
  但是这个 $72$ 仍大于我们在圆心附近得到的 $\tfrac{200}{3}\approx 66.7$ 。因此圆内的临界点才给出真正全局最小值。

结论：

最小值 出现在内点 $\bigl(x,y\bigr)=\bigl(2+\tfrac{2}{3},\,2\bigr)$ （即 $\tfrac{2}{3}$ 朝 $x$ -方向偏移），其值
$\min\bigl(PA^2 + PB^2 + PC^2\bigr) \;=\; \frac{200}{3}.$
最大值 出现在内切圆左端点 $\bigl(x,y\bigr) = (0,2)$ ，其值
$\max\bigl(PA^2 + PB^2 + PC^2\bigr) \;=\; 88.$

答案

\boxed{ \text{最小值 } = \; \frac{200}{3}, \quad \text{最大值 } = \; 88. }

Frage

Gedanke

Antwort

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