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rotvap 写了： 2025年 7月 17日 17:39

半圆左边要离开A一点点

huangchong 写了： 2025年 7月 17日 17:56 半圆左边要离开A一点点

没想到还有这个操作

huangchong 写了： 2025年 7月 17日 13:06 因为顶上没碰着 把这个半圆一端往上偏转 可以得一个更大的半圆 另一端当然得往后缩一点 这题要仔细分析其实是比较麻烦的 但是可以假设有个半圆 一个直径靠两边 圆弧切另两边 这个形状可以想象 半径大小也不难算（高中内容）

高是 4厘米就对了， 笔误

你初中高中应该班级成绩排名不高。

rotvap 写了： 2025年 7月 17日 17:48 我觉得大家想多了

底边这样一转，必然会有一段圆弧跑到矩形外面去

必须是8的边当直径

这样两条6的边才能是切线，保证半圆不会跑出去

其实正解应该就是老师想的简单答案

要把8和6作为变量，来考察是否同时相切才是周长极值。

量子战士 写了： 2025年 7月 17日 17:42 你这肯定是错的。

见下图



半径R与角a就按蝗总的方程走。

我草草算了一下，半径是13/3厘米。不过，我列了4个方程，很可能中间有运算错误。

huangchong 写了： 2025年 7月 17日 12:21 应该是出题的人自己没想到把半圆从底边往上转还可以更大

似乎是（pi+2）（14-4sqrt（6））

我算了下，是14-4倍根6.。。。。。 x的平方-28x+100=0 的其中一个实数解。

如果用deepseek 算出来是3分之13， 但这个结果代入的话半圆理论上会出长方形。

量子战士 写了： 2025年 7月 17日 17:42 你这肯定是错的。

见下图



半径R与角a就按蝗总的方程走。

R+R*cos(theta) = 8
R*sin(theta)+R = 6

solve for R and theta




To solve the system of equations:
R + R*cos(theta) = 8
R*sin(theta) + R = 6
First, factor out R from both equations:
R(1 + cos(theta)) = 8
R(1 + sin(theta)) = 6
From these, we can express R in terms of theta:
From (1): R = 8 / (1 + cos(theta))
From (2): R = 6 / (1 + sin(theta))
Since both expressions equal R, we can set them equal to each other:
8 / (1 + cos(theta)) = 6 / (1 + sin(theta))
Cross-multiply to eliminate the denominators:
8(1 + sin(theta)) = 6(1 + cos(theta))
8 + 8sin(theta) = 6 + 6cos(theta)
Rearrange the terms to get a trigonometric equation:
2 + 8sin(theta) = 6cos(theta)
Divide by 2:
1 + 4sin(theta) = 3cos(theta)
To solve this trigonometric equation, we use the tangent half-angle substitution, where t = tan(theta/2).
Recall the identities: sin(theta) = 2t / (1 + t^2) and cos(theta) = (1 - t^2) / (1 + t^2).
Substitute these into the equation:
1 + 4 * [2t / (1 + t^2)] = 3 * [(1 - t^2) / (1 + t^2)]
1 + 8t / (1 + t^2) = 3(1 - t^2) / (1 + t^2)
Multiply the entire equation by (1 + t^2) to clear the denominators:
(1 + t^2) + 8t = 3(1 - t^2)
1 + t^2 + 8t = 3 - 3t^2
Rearrange into a quadratic equation in terms of t:
t^2 + 3t^2 + 8t + 1 - 3 = 0
4t^2 + 8t - 2 = 0
Divide by 2:
2t^2 + 4t - 1 = 0
Use the quadratic formula t = [-b ± sqrt(b^2 - 4ac)] / 2a:
t = [-4 ± sqrt(4^2 - 4 * 2 * -1)] / (2 * 2)
t = [-4 ± sqrt(16 + 8)] / 4
t = [-4 ± sqrt(24)] / 4
t = [-4 ± 2*sqrt(6)] / 4
t = -1 ± sqrt(6)/2
So we have two possible values for t = tan(theta/2):
t1 = -1 + sqrt(6)/2
t2 = -1 - sqrt(6)/2
Now, we find theta from t = tan(theta/2), so theta = 2 * arctan(t).
Solution 1: Using t1 = -1 + sqrt(6)/2
theta1/2 = arctan(-1 + sqrt(6)/2)
theta1 = 2 * arctan(-1 + sqrt(6)/2)
Numerically:
sqrt(6) ≈ 2.44949
t1 ≈ -1 + 2.44949/2 ≈ -1 + 1.224745 ≈ 0.224745
theta1/2 ≈ arctan(0.224745) ≈ 0.22129 radians ≈ 12.68 degrees
theta1 ≈ 2 * 0.22129 radians ≈ 0.44258 radians ≈ 25.36 degrees
Now, find R using R = 8 / (1 + cos(theta1)) or R = 6 / (1 + sin(theta1)).
It's easier to use the exact values of cos(theta) and sin(theta) from t:
cos(theta1) = (1 - t1^2) / (1 + t1^2) and sin(theta1) = 2t1 / (1 + t1^2)
After algebraic simplification (as shown in thought process), we find:
cos(theta1) = (3 + 8sqrt(6)) / 25
sin(theta1) = (-4 + 6sqrt(6)) / 25
Substitute into R = 8 / (1 + cos(theta1)):
R1 = 8 / (1 + (3 + 8sqrt(6))/25) = 8 / ((25 + 3 + 8sqrt(6))/25) = 8 * 25 / (28 + 8sqrt(6))
R1 = 200 / (28 + 8sqrt(6)) = 50 / (7 + 2sqrt(6))
Rationalize the denominator:
R1 = 50 * (7 - 2sqrt(6)) / ((7 + 2sqrt(6))(7 - 2sqrt(6))) = 50 * (7 - 2sqrt(6)) / (49 - 4*6)
R1 = 50 * (7 - 2sqrt(6)) / (49 - 24) = 50 * (7 - 2sqrt(6)) / 25
R1 = 2 * (7 - 2sqrt(6)) = 14 - 4sqrt(6)
Numerically: R1 ≈ 14 - 4 * 2.44949 ≈ 14 - 9.79796 ≈ 4.20204
Solution 2: Using t2 = -1 - sqrt(6)/2
theta2/2 = arctan(-1 - sqrt(6)/2)
theta2 = 2 * arctan(-1 - sqrt(6)/2)
Numerically:
t2 ≈ -1 - 1.224745 ≈ -2.224745
theta2/2 ≈ arctan(-2.224745) ≈ -1.14799 radians ≈ -65.76 degrees
theta2 ≈ 2 * -1.14799 radians ≈ -2.29598 radians ≈ -131.52 degrees
To express in the range [0, 2pi), add 2pi:
theta2 ≈ -2.29598 + 2pi ≈ 3.9872 radians ≈ 228.48 degrees
Now, find R using R = 8 / (1 + cos(theta2)) or R = 6 / (1 + sin(theta2)).
cos(theta2) = (1 - t2^2) / (1 + t2^2) and sin(theta2) = 2t2 / (1 + t2^2)
After algebraic simplification, we find:
cos(theta2) = (3 - 8sqrt(6)) / 25
sin(theta2) = (-4 - 6sqrt(6)) / 25
Substitute into R = 8 / (1 + cos(theta2)):
R2 = 8 / (1 + (3 - 8sqrt(6))/25) = 8 / ((25 + 3 - 8sqrt(6))/25) = 8 * 25 / (28 - 8sqrt(6))
R2 = 200 / (28 - 8sqrt(6)) = 50 / (7 - 2sqrt(6))
Rationalize the denominator:
R2 = 50 * (7 + 2sqrt(6)) / ((7 - 2sqrt(6))(7 + 2sqrt(6))) = 50 * (7 + 2sqrt(6)) / (49 - 4*6)
R2 = 50 * (7 + 2sqrt(6)) / (49 - 24) = 50 * (7 + 2sqrt(6)) / 25
R2 = 2 * (7 + 2sqrt(6)) = 14 + 4sqrt(6)
Numerically: R2 ≈ 14 + 4 * 2.44949 ≈ 14 + 9.79796 ≈ 23.79796
Summary of Solutions:
There are two pairs of solutions for (R, theta):
Solution 1:
R = 14 - 4sqrt(6) (approximately 4.202)
theta = 2 * arctan(-1 + sqrt(6)/2) (approximately 0.4426 radians or 25.36 degrees)
Solution 2:
R = 14 + 4sqrt(6) (approximately 23.798)
theta = 2 * arctan(-1 - sqrt(6)/2) (approximately 3.9872 radians or 228.48 degrees)
Note: For the angle theta, general solutions would include adding multiples of 2π (or 360 degrees) due to the periodic nature of trigonometric functions. The provided theta values are within the range [0, 2π).).

你们把小学数学题做成了研究生题目了。

glowinglake 写了： 2025年 7月 18日 01:12 R+R*cos(theta) = 8
R*sin(theta)+R = 6

solve for R and theta




To solve the system of equations:
R + R*cos(theta) = 8
R*sin(theta) + R = 6
First, factor out R from both equations:
R(1 + cos(theta)) = 8
R(1 + sin(theta)) = 6
From these, we can express R in terms of theta:
From (1): R = 8 / (1 + cos(theta))
From (2): R = 6 / (1 + sin(theta))
Since both expressions equal R, we can set them equal to each other:
8 / (1 + cos(theta)) = 6 / (1 + sin(theta))
Cross-multiply to eliminate the denominators:
8(1 + sin(theta)) = 6(1 + cos(theta))
8 + 8sin(theta) = 6 + 6cos(theta)
Rearrange the terms to get a trigonometric equation:
2 + 8sin(theta) = 6cos(theta)
Divide by 2:
1 + 4sin(theta) = 3cos(theta)
To solve this trigonometric equation, we use the tangent half-angle substitution, where t = tan(theta/2).
Recall the identities: sin(theta) = 2t / (1 + t^2) and cos(theta) = (1 - t^2) / (1 + t^2).
Substitute these into the equation:
1 + 4 * [2t / (1 + t^2)] = 3 * [(1 - t^2) / (1 + t^2)]
1 + 8t / (1 + t^2) = 3(1 - t^2) / (1 + t^2)
Multiply the entire equation by (1 + t^2) to clear the denominators:
(1 + t^2) + 8t = 3(1 - t^2)
1 + t^2 + 8t = 3 - 3t^2
Rearrange into a quadratic equation in terms of t:
t^2 + 3t^2 + 8t + 1 - 3 = 0
4t^2 + 8t - 2 = 0
Divide by 2:
2t^2 + 4t - 1 = 0
Use the quadratic formula t = [-b ± sqrt(b^2 - 4ac)] / 2a:
t = [-4 ± sqrt(4^2 - 4 * 2 * -1)] / (2 * 2)
t = [-4 ± sqrt(16 + 8)] / 4
t = [-4 ± sqrt(24)] / 4
t = [-4 ± 2*sqrt(6)] / 4
t = -1 ± sqrt(6)/2
So we have two possible values for t = tan(theta/2):
t1 = -1 + sqrt(6)/2
t2 = -1 - sqrt(6)/2
Now, we find theta from t = tan(theta/2), so theta = 2 * arctan(t).
Solution 1: Using t1 = -1 + sqrt(6)/2
theta1/2 = arctan(-1 + sqrt(6)/2)
theta1 = 2 * arctan(-1 + sqrt(6)/2)
Numerically:
sqrt(6) ≈ 2.44949
t1 ≈ -1 + 2.44949/2 ≈ -1 + 1.224745 ≈ 0.224745
theta1/2 ≈ arctan(0.224745) ≈ 0.22129 radians ≈ 12.68 degrees
theta1 ≈ 2 * 0.22129 radians ≈ 0.44258 radians ≈ 25.36 degrees
Now, find R using R = 8 / (1 + cos(theta1)) or R = 6 / (1 + sin(theta1)).
It's easier to use the exact values of cos(theta) and sin(theta) from t:
cos(theta1) = (1 - t1^2) / (1 + t1^2) and sin(theta1) = 2t1 / (1 + t1^2)
After algebraic simplification (as shown in thought process), we find:
cos(theta1) = (3 + 8sqrt(6)) / 25
sin(theta1) = (-4 + 6sqrt(6)) / 25
Substitute into R = 8 / (1 + cos(theta1)):
R1 = 8 / (1 + (3 + 8sqrt(6))/25) = 8 / ((25 + 3 + 8sqrt(6))/25) = 8 * 25 / (28 + 8sqrt(6))
R1 = 200 / (28 + 8sqrt(6)) = 50 / (7 + 2sqrt(6))
Rationalize the denominator:
R1 = 50 * (7 - 2sqrt(6)) / ((7 + 2sqrt(6))(7 - 2sqrt(6))) = 50 * (7 - 2sqrt(6)) / (49 - 4*6)
R1 = 50 * (7 - 2sqrt(6)) / (49 - 24) = 50 * (7 - 2sqrt(6)) / 25
R1 = 2 * (7 - 2sqrt(6)) = 14 - 4sqrt(6)
Numerically: R1 ≈ 14 - 4 * 2.44949 ≈ 14 - 9.79796 ≈ 4.20204
Solution 2: Using t2 = -1 - sqrt(6)/2
theta2/2 = arctan(-1 - sqrt(6)/2)
theta2 = 2 * arctan(-1 - sqrt(6)/2)
Numerically:
t2 ≈ -1 - 1.224745 ≈ -2.224745
theta2/2 ≈ arctan(-2.224745) ≈ -1.14799 radians ≈ -65.76 degrees
theta2 ≈ 2 * -1.14799 radians ≈ -2.29598 radians ≈ -131.52 degrees
To express in the range [0, 2pi), add 2pi:
theta2 ≈ -2.29598 + 2pi ≈ 3.9872 radians ≈ 228.48 degrees
Now, find R using R = 8 / (1 + cos(theta2)) or R = 6 / (1 + sin(theta2)).
cos(theta2) = (1 - t2^2) / (1 + t2^2) and sin(theta2) = 2t2 / (1 + t2^2)
After algebraic simplification, we find:
cos(theta2) = (3 - 8sqrt(6)) / 25
sin(theta2) = (-4 - 6sqrt(6)) / 25
Substitute into R = 8 / (1 + cos(theta2)):
R2 = 8 / (1 + (3 - 8sqrt(6))/25) = 8 / ((25 + 3 - 8sqrt(6))/25) = 8 * 25 / (28 - 8sqrt(6))
R2 = 200 / (28 - 8sqrt(6)) = 50 / (7 - 2sqrt(6))
Rationalize the denominator:
R2 = 50 * (7 + 2sqrt(6)) / ((7 - 2sqrt(6))(7 + 2sqrt(6))) = 50 * (7 + 2sqrt(6)) / (49 - 4*6)
R2 = 50 * (7 + 2sqrt(6)) / (49 - 24) = 50 * (7 + 2sqrt(6)) / 25
R2 = 2 * (7 + 2sqrt(6)) = 14 + 4sqrt(6)
Numerically: R2 ≈ 14 + 4 * 2.44949 ≈ 14 + 9.79796 ≈ 23.79796
Summary of Solutions:
There are two pairs of solutions for (R, theta):
Solution 1:
R = 14 - 4sqrt(6) (approximately 4.202)
theta = 2 * arctan(-1 + sqrt(6)/2) (approximately 0.4426 radians or 25.36 degrees)
Solution 2:
R = 14 + 4sqrt(6) (approximately 23.798)
theta = 2 * arctan(-1 - sqrt(6)/2) (approximately 3.9872 radians or 228.48 degrees)
Note: For the angle theta, general solutions would include adding multiples of 2π (or 360 degrees) due to the periodic nature of trigonometric functions. The provided theta values are within the range [0, 2π).).

ai?真能折腾

果真是术版

同一个链接里有人用订阅的chatgpt4o居然能够直接得到一般解？尝试了好像没有哪个大模型能够直接做对

rotvap 写了： 2025年 7月 17日 17:39

不对吧，直径的端点一定在A吗？

小学知识够解这题，只要用勾股定理加中位线定理就够了，哪有这么复杂。

zheliemit 写了： 2025年 7月 18日 01:34 你们把小学数学题做成了研究生题目了。

x+r=8
y+r=6
x^2+y^2=r^2
很简单就能解出r的值。

14 非工程师也一概而过

zeami 写了： 2025年 7月 17日 17:49 还在算？我们攻城狮已经一分钟算好了哇。理论直径在8～10之间，目测在9以下，然后取整就是14。。