# Thread:

1. ## Proof for a space geometry question

Show that there exist one point equidistant from four non complainer distinct points (maybe it can be solved using axis of a circle )

2. ## Re: Proof for a space geometry question

Originally Posted by lebanon
Show that there exist one point equidistant from four non complainer distinct points (maybe it can be solved using axis of a circle )
1. Four non-complanar points are located on a sphere. Thus you are looking for it's center.

2. If the center of the sphere is $C(c_1, c_2, c_3)$ then the equation of the sphere is:

$(x-c_1)^2+(y-c_2)^2+(z-c_3)^2=r^2$

Plug in the coordinates of the four points. You'll get a system of 4 equations which can be easily reduced to a system of 3 linear equations. Solve for $(c_1, c_2, c_3)$.
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 3, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9125584363937378, "perplexity_flag": "head"}