上面有幅grok4 solve咗㗎
唔通係grok 4 heavy?
史上最難小學五年級數學題目
紫式ぼう
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上面有幅grok4 solve咗㗎
唔通係grok 4 heavy?
小5數應該唔使拆三角形嘅，不過出題果個咁樣出真係會搞左啲奇怪野出黎
數學體育老師教？
你笨柒弱智就咪撚當我On9囉！！
你仲識唔識用算盤快算7位數？仲識唔識用4-FigureTable？
你仲識唔識用算盤快算7位數？仲識唔識用4-FigureTable？
得，不過唔係最大囉
緊係唔係啦, 我當年個數學老師好厲害的, 佢除左教數學都有教美術的.
你user name好襯你 
呢個方法都應該係答案
但係有冇方法嚴謹啲證明到佢就係最大
但係有冇方法嚴謹啲證明到佢就係最大
咁你應該要用微積分了
衝出嚟柒就係你依啲戇鳩 
呢類搵最大既值都好撚難
有好多數學未解決既難題都係 某人猜想
跟著證明最大既值係咩
有好多數學未解決既難題都係 某人猜想
跟著證明最大既值係咩
痴撚線題目其實本意係想人拎邊個做答案？
其實本意係想人拎邊個做答案？係咪無 deepseek
難撚到
The formula arises from solving the geometric relationships for an inscribed semicircle:
1. Setup: Consider a rectangle from (0,0) to (m,n). Place one endpoint of the diameter at (x,0) on the bottom side and the other at (0,y) on the left side.
• The center of the semicircle is at (x/2, y/2).
• The radius r satisfies the diameter length: 2r = √(x² + y²).
2. Tangency Constraints: For the arc to be tangent to the right side (x = m) and top side (y = n):
• Distance from center to right side: m - x/2 = r → x = 2(m - r).
• Distance from center to top side: n - y/2 = r → y = 2(n - r).
3. Substitute and Solve:
• Plug into the diameter equation: 2r = √(2(m - r))² + (2(n - r))².
• Simplify: r = √(m - r)² + (n - r)².
• Square both sides: r² = (m - r)² + (n - r)².
• Expand: r² = m² - 2mr + r² + n² - 2nr + r².
• Rearrange: 0 = r² + m² + n² - 2r(m + n).
• Quadratic form: r² - 2(m + n)r + (m² + n²) = 0.
• Solve using quadratic formula: r = (m + n) ± √(m + n)² - (m² + n²) = (m + n) ± √(2mn).
• Select the smaller root (minus sign) for physical feasibility: r = (m + n) - √(2mn).
The formula arises from solving the geometric relationships for an inscribed semicircle:
1. Setup: Consider a rectangle from (0,0) to (m,n). Place one endpoint of the diameter at (x,0) on the bottom side and the other at (0,y) on the left side.
• The center of the semicircle is at (x/2, y/2).
• The radius r satisfies the diameter length: 2r = √(x² + y²).
2. Tangency Constraints: For the arc to be tangent to the right side (x = m) and top side (y = n):
• Distance from center to right side: m - x/2 = r → x = 2(m - r).
• Distance from center to top side: n - y/2 = r → y = 2(n - r).
3. Substitute and Solve:
• Plug into the diameter equation: 2r = √(2(m - r))² + (2(n - r))².
• Simplify: r = √(m - r)² + (n - r)².
• Square both sides: r² = (m - r)² + (n - r)².
• Expand: r² = m² - 2mr + r² + n² - 2nr + r².
• Rearrange: 0 = r² + m² + n² - 2r(m + n).
• Quadratic form: r² - 2(m + n)r + (m² + n²) = 0.
• Solve using quadratic formula: r = (m + n) ± √(m + n)² - (m² + n²) = (m + n) ± √(2mn).
• Select the smaller root (minus sign) for physical feasibility: r = (m + n) - √(2mn).
25.12厘米
如果個長方形係4x8cm就可以變返做小5
題
題
咁難先3分？？？？？？
用quadratic equation計完個r都仲未計面積同 circumference
中五都未必識答
又會咁啱同po條題目一樣比例？似係例題，如果係咁就明顯唔係考緊數學能力，係考操卷同背答案能力
咁哩位人兄又係教人答8cm?
試左幾個AI，第一次答都係8cm
要再比提示叫佢打斜放先識
要再比提示叫佢打斜放先識
因為好多人都唔熟circle theorems