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Home » \[ x = 80 \, \textmL \] - Baxtercollege

\[ x = 80 \, \textmL \] - Baxtercollege

📅 March 10, 2026 👤 scraface

Mar 10, 2026

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📌 الحل: لنجعل رؤوس الهرم الرباعي المنتظم هي $A = (0,0,0)$ و $B = (a,a,0)$ و $C = (0,a,a)$. لإيجاد إحداثيات الرأس الرابع $D = (x,y,z)$، نستخدم خصائص الهرم الرباعي المنتظم، حيث تكون جميع الحواف متساوية في الطول. طول الحافة $s$ يُعطى بالمسافة بين $A$ و $B$:

📌 s = \sqrt{(a-0)^2 + (a-0)^2 + (0-0)^2} = \sqrt{2a^2} = a\sqrt{2}

📌 نحن بحاجة إلى أن تكون المسافات من $D$ إلى $A$ و $B$ و $C$ جميعها تساوي $a\sqrt{2}$. أولاً، نحسب المسافة من $D$ إلى $A$:

📌 \sqrt{x^2 + y^2 + z^2} = a\sqrt{2} \quad \Rightarrow \quad x^2 + y^2 + z^2 = 2a^2

📌 بعد ذلك، المسافة من $D$ إلى $B$:

📌 \sqrt{(x-a)^2 + (y-a)^2 + z^2} = a\sqrt{2} \quad \Rightarrow \quad (x-a)^2 + (y-a)^2 + z^2 = 2a^2

📌 بتوسيع وتعويض $x^2 + y^2 + z^2 = 2a^2$:

📌 x^2 - 2ax + a^2 + y^2 - 2ay + a^2 + z^2 = 2a^2 \quad \Rightarrow \quad 2a^2 - 2ax - 2ay + 2a^2 = 2a^2 \quad \Rightarrow \quad 2a^2 - 2ax - 2ay + 2a^2 - 2a^2 = 0 \quad \Rightarrow \quad ax + ay = a^2 \quad \Rightarrow \quad x + y = a

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