https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=SnowStorm&feedformat=atom AoPS Wiki - User contributions [en] 2020-11-29T05:14:09Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=User_talk:SnowStorm&diff=3496 User talk:SnowStorm 2006-06-20T21:03:13Z <p>SnowStorm: </p> <hr /> <div>Per Mr. Crawford's suggestion, I will keep a list of my contributions.<br /> &lt;br&gt;&lt;br&gt;<br /> 1) [[Wilson's Theorem]]&lt;br&gt;<br /> 2) Sample of [[Fermat's Little Theorem]]&lt;br&gt;<br /> 3) [[Divisibility]]&lt;br&gt;<br /> 4) [[Ceva's Theorem]]&lt;br&gt;<br /> 5) [[Menelaus' Theorem]]&lt;br&gt;<br /> 6) [[Stewart's Theorem]]&lt;br&gt;<br /> 7) [[Overcounting]]&lt;br&gt;<br /> 8) [[Cevian]]&lt;br&gt;<br /> 9) [[Median]]&lt;br&gt;<br /> 10) [[Altitude]]&lt;br&gt;<br /> 11) [[Angle Bisector]]&lt;br&gt;<br /> <br /> TODO&lt;br&gt;<br /> 4) &lt;strike&gt;Add proof&lt;/strike&gt;, &lt;strike&gt;clean up example&lt;/strike&gt;&lt;br&gt;<br /> 5) Add proof, add example&lt;br&gt;<br /> 6) &lt;strike&gt;Add proof&lt;/strike&gt;, add example&lt;br&gt;<br /> 7) &lt;strike&gt;Find better example&lt;/strike&gt;</div> SnowStorm https://artofproblemsolving.com/wiki/index.php?title=Ceva%27s_Theorem&diff=3494 Ceva's Theorem 2006-06-20T21:02:40Z <p>SnowStorm: /* Proof */</p> <hr /> <div>'''Ceva's Theorem''' is an algebraic statement regarding the lengths of [[Cevian|cevians]] in a [[triangle]].<br /> <br /> <br /> == Statement ==<br /> ''(awaiting image)''&lt;br&gt;<br /> A necessary and sufficient condition for AD, BE, CF, where D, E, and F are points of the respective side lines BC, CA, AB of a triangle ABC, to be concurrent is that<br /> &lt;br&gt;&lt;center&gt;&lt;math&gt;BD * CE * AF = +DC * EA * FB&lt;/math&gt;&lt;/center&gt;&lt;br&gt;<br /> where all segments in the formula are directed segments.<br /> <br /> == Proof ==<br /> Let &lt;math&gt;{X,Y,Z}&lt;/math&gt; be points on &lt;math&gt;{BC}, {CA}, {AB}&lt;/math&gt; respectively such that &lt;math&gt;AX,BY,CZ&lt;/math&gt; are concurrent, and let &lt;math&gt;{P}&lt;/math&gt; be the point where &lt;math&gt;AX&lt;/math&gt;, &lt;math&gt;BY&lt;/math&gt; and &lt;math&gt;CZ&lt;/math&gt; meet. Draw a parallel to &lt;math&gt;AB&lt;/math&gt; through the point &lt;math&gt;{C}&lt;/math&gt;. Extend &lt;math&gt;AX&lt;/math&gt; until it intersects the parallel at a point &lt;math&gt;\displaystyle{A'}&lt;/math&gt;. Construct &lt;math&gt;\displaystyle{B'}&lt;/math&gt; in a similar way extending &lt;math&gt;BY&lt;/math&gt;.<br /> &lt;center&gt;''(ceva1.png)''&lt;/center&gt;<br /> The triangles &lt;math&gt;\displaystyle{\triangle{ABX}}&lt;/math&gt; and &lt;math&gt;\displaystyle{\triangle{A'CX}}&lt;/math&gt; are similar, and so are &lt;math&gt;\displaystyle\triangle{ABY}&lt;/math&gt; and &lt;math&gt;\triangle{CB'Y}&lt;/math&gt;. Then the following equalities hold:<br /> &lt;center&gt;&lt;math&gt;\frac{BX}{XC}=\frac{AB}{CA'},\qquad\frac{CY}{YA}=\frac{CB'}{BA}&lt;/math&gt;&lt;/center&gt;<br /> &lt;br&gt;<br /> and thus<br /> &lt;center&gt;&lt;math&gt;\frac{BX}{XC}\cdot\frac{CY}{YA}=\frac{AB}{CA'}\cdot\frac{CB'}{BA}=\frac{CB'}{A'C} \qquad(1)&lt;/math&gt;&lt;/center&gt;<br /> &lt;br&gt;<br /> Notice that if directed segments are being used then &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;BA&lt;/math&gt; have opposite signs, and therefore when cancelled change the sign of the expression. That's why we changed &lt;math&gt;CA'&lt;/math&gt; to &lt;math&gt;A'C&lt;/math&gt;.<br /> &lt;br&gt;&lt;br&gt;<br /> Now we turn to consider the following similarities: &lt;math&gt;\triangle{AZP}\sim\triangle{A'CP}&lt;/math&gt; and &lt;math&gt;\triangle BZP\sim\triangle B'CP&lt;/math&gt;. From them we get the equalities<br /> &lt;center&gt;&lt;math&gt;\frac{CP}{ZP}=\frac{A'C}{AZ},\qquad\frac{CP}{ZP}=\frac{CB'}{ZB}&lt;/math&gt;&lt;/center&gt;<br /> &lt;br&gt;<br /> which lead to<br /> &lt;center&gt;&lt;math&gt;\frac{AZ}{ZB}=\frac{A'C}{CB'}&lt;/math&gt;.&lt;/center&gt;<br /> &lt;br&gt;<br /> Multiplying the last expression with (1) gives<br /> &lt;center&gt;&lt;math&gt;\frac{AZ}{ZB}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1&lt;/math&gt;&lt;/center&gt;<br /> &lt;br&gt;<br /> and we conclude the proof.<br /> &lt;br&gt;&lt;br&gt;<br /> To prove the converse, suppose that &lt;math&gt;{X,Y,Z}&lt;/math&gt; are points on &lt;math&gt;{BC}, {CA}, {AB}&lt;/math&gt; respectively and satisfying<br /> &lt;center&gt;&lt;math&gt;\frac{AZ}{ZB}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1.&lt;/math&gt;&lt;/center&gt;<br /> &lt;br&gt;<br /> Let &lt;math&gt;Q&lt;/math&gt; be the intersection point of &lt;math&gt;AX&lt;/math&gt; with &lt;math&gt;BY&lt;/math&gt;, and let &lt;math&gt;Z'&lt;/math&gt; be the intersection of &lt;math&gt;CQ&lt;/math&gt; with &lt;math&gt;AB&lt;/math&gt;. Since then &lt;math&gt;AX,BY,CZ'&lt;/math&gt; are concurrent, we have<br /> &lt;center&gt;&lt;math&gt;\frac{AZ'}{Z'B}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1&lt;/math&gt;&lt;/center&gt;<br /> &lt;br&gt;<br /> and thus<br /> &lt;center&gt;&lt;math&gt;\frac{AZ'}{Z'B}=\frac{AZ}{ZB}&lt;/math&gt;&lt;/center&gt;<br /> &lt;br&gt;<br /> which implies &lt;math&gt;Z=Z'&lt;/math&gt;, and therefore &lt;math&gt;AX,BY,CZ&lt;/math&gt; are concurrent.<br /> <br /> &lt;div align=&quot;right&quot;&gt;''(proof courtesy planetmath.org, used under GNU License)''&lt;/div&gt;<br /> <br /> == Example ==<br /> Suppose AB, AC, and BC have lengths 13, 14, and 15. If &lt;math&gt;\frac{AF}{FB} = \frac{2}{5}&lt;/math&gt; and &lt;math&gt;\frac{CE}{EA} = \frac{5}{8}&lt;/math&gt;. Find BD and DC.&lt;br&gt;<br /> &lt;br&gt;<br /> If &lt;math&gt;BD = x&lt;/math&gt; and &lt;math&gt;DC = y&lt;/math&gt;, then &lt;math&gt;10x = 40y&lt;/math&gt;, and &lt;math&gt;{x + y = 15}&lt;/math&gt;. From this, we find &lt;math&gt;x = 12&lt;/math&gt; and &lt;math&gt;y = 3&lt;/math&gt;.<br /> <br /> == See also ==<br /> * [[Menelaus' Theorem]]<br /> * [[Stewart's Theorem]]</div> SnowStorm https://artofproblemsolving.com/wiki/index.php?title=Ceva%27s_Theorem&diff=3493 Ceva's Theorem 2006-06-20T21:01:26Z <p>SnowStorm: /* Proof */</p> <hr /> <div>'''Ceva's Theorem''' is an algebraic statement regarding the lengths of [[Cevian|cevians]] in a [[triangle]].<br /> <br /> <br /> == Statement ==<br /> ''(awaiting image)''&lt;br&gt;<br /> A necessary and sufficient condition for AD, BE, CF, where D, E, and F are points of the respective side lines BC, CA, AB of a triangle ABC, to be concurrent is that<br /> &lt;br&gt;&lt;center&gt;&lt;math&gt;BD * CE * AF = +DC * EA * FB&lt;/math&gt;&lt;/center&gt;&lt;br&gt;<br /> where all segments in the formula are directed segments.<br /> <br /> == Proof ==<br /> Let &lt;math&gt;{X,Y,Z}&lt;/math&gt; be points on &lt;math&gt;{BC}, {CA}, {AB}&lt;/math&gt; respectively such that &lt;math&gt;AX,BY,CZ&lt;/math&gt; are concurrent, and let &lt;math&gt;{P}&lt;/math&gt; be the point where &lt;math&gt;AX&lt;/math&gt;, &lt;math&gt;BY&lt;/math&gt; and &lt;math&gt;CZ&lt;/math&gt; meet. Draw a parallel to &lt;math&gt;AB&lt;/math&gt; through the point &lt;math&gt;{C}&lt;/math&gt;. Extend &lt;math&gt;AX&lt;/math&gt; until it intersects the parallel at a point &lt;math&gt;\displaystyle{A'}&lt;/math&gt;. Construct &lt;math&gt;\displaystyle{B'}&lt;/math&gt; in a similar way extending &lt;math&gt;BY&lt;/math&gt;.<br /> &lt;center&gt;''(ceva1.png)''&lt;/center&gt;<br /> The triangles &lt;math&gt;\displaystyle{\triangle{ABX}}&lt;/math&gt; and &lt;math&gt;\displaystyle{\triangle{A'CX}}&lt;/math&gt; are similar, and so are &lt;math&gt;\displaystyle\triangle{ABY}&lt;/math&gt; and &lt;math&gt;\triangle{CB'Y}&lt;/math&gt;. Then the following equalities hold:<br /> &lt;center&gt;&lt;math&gt;\frac{BX}{XC}=\frac{AB}{CA'},\qquad\frac{CY}{YA}=\frac{CB'}{BA}&lt;/math&gt;&lt;/center&gt;<br /> &lt;br&gt;<br /> and thus<br /> &lt;center&gt;&lt;math&gt;\frac{BX}{XC}\cdot\frac{CY}{YA}=\frac{AB}{CA'}\cdot\frac{CB'}{BA}=\frac{CB'}{A'C} \qquad(1)&lt;/math&gt;&lt;/center&gt;<br /> &lt;br&gt;<br /> Notice that if directed segments are being used then &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;BA&lt;/math&gt; have opposite signs, and therefore when cancelled change the sign of the expression. That's why we changed &lt;math&gt;CA'&lt;/math&gt; to &lt;math&gt;A'C&lt;/math&gt;.<br /> &lt;br&gt;&lt;br&gt;<br /> Now we turn to consider the following similarities: &lt;math&gt;\triangle{AZP}\sim\triangle{A'CP}&lt;/math&gt; and &lt;math&gt;\triangle BZP\sim\triangle B'CP&lt;/math&gt;. From them we get the equalities<br /> &lt;center&gt;&lt;math&gt;\frac{CP}{ZP}=\frac{A'C}{AZ},\qquad\frac{CP}{ZP}=\frac{CB'}{ZB}&lt;/math&gt;&lt;/center&gt;<br /> &lt;br&gt;<br /> which lead to<br /> &lt;center&gt;&lt;math&gt;\frac{AZ}{ZB}=\frac{A'C}{CB'}&lt;/math&gt;.&lt;/center&gt;<br /> &lt;br&gt;<br /> Multiplying the last expression with (1) gives<br /> &lt;center&gt;&lt;math&gt;\frac{AZ}{ZB}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1&lt;/math&gt;&lt;/center&gt;<br /> &lt;br&gt;<br /> and we conclude the proof.<br /> &lt;br&gt;&lt;br&gt;<br /> To prove the converse, suppose that &lt;math&gt;{X,Y,Z}&lt;/math&gt; are points on &lt;math&gt;{BC}, {CA}, {AB}&lt;/math&gt; respectively and satisfying<br /> &lt;center&gt;&lt;math&gt;\frac{AZ}{ZB}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1.&lt;/math&gt;&lt;/center&gt;<br /> &lt;br&gt;<br /> Let &lt;math&gt;Q&lt;/math&gt; be the intersection point of &lt;math&gt;AX&lt;/math&gt; with &lt;math&gt;BY&lt;/math&gt;, and let &lt;math&gt;Z'&lt;/math&gt; be the intersection of &lt;math&gt;CQ&lt;/math&gt; with &lt;math&gt;AB&lt;/math&gt;. Since then &lt;math&gt;AX,BY,CZ'&lt;/math&gt; are concurrent, we have<br /> &lt;center&gt;&lt;math&gt;\frac{AZ'}{Z'B}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1&lt;/math&gt;&lt;/center&gt;<br /> &lt;br&gt;<br /> and thus<br /> &lt;center&gt;&lt;math&gt;\frac{AZ'}{Z'B}=\frac{AZ}{ZB}&lt;/math&gt;&lt;/center&gt;<br /> &lt;br&gt;<br /> which implies &lt;math&gt;Z=Z'&lt;/math&gt;, and therefore &lt;math&gt;AX,BY,CZ&lt;/math&gt; are concurrent.<br /> <br /> == Example ==<br /> Suppose AB, AC, and BC have lengths 13, 14, and 15. If &lt;math&gt;\frac{AF}{FB} = \frac{2}{5}&lt;/math&gt; and &lt;math&gt;\frac{CE}{EA} = \frac{5}{8}&lt;/math&gt;. Find BD and DC.&lt;br&gt;<br /> &lt;br&gt;<br /> If &lt;math&gt;BD = x&lt;/math&gt; and &lt;math&gt;DC = y&lt;/math&gt;, then &lt;math&gt;10x = 40y&lt;/math&gt;, and &lt;math&gt;{x + y = 15}&lt;/math&gt;. From this, we find &lt;math&gt;x = 12&lt;/math&gt; and &lt;math&gt;y = 3&lt;/math&gt;.<br /> <br /> == See also ==<br /> * [[Menelaus' Theorem]]<br /> * [[Stewart's Theorem]]</div> SnowStorm https://artofproblemsolving.com/wiki/index.php?title=Ceva%27s_Theorem&diff=3485 Ceva's Theorem 2006-06-20T20:38:35Z <p>SnowStorm: /* Proof */</p> <hr /> <div>'''Ceva's Theorem''' is an algebraic statement regarding the lengths of [[Cevian|cevians]] in a [[triangle]].<br /> <br /> <br /> == Statement ==<br /> ''(awaiting image)''&lt;br&gt;<br /> A necessary and sufficient condition for AD, BE, CF, where D, E, and F are points of the respective side lines BC, CA, AB of a triangle ABC, to be concurrent is that<br /> &lt;br&gt;&lt;center&gt;&lt;math&gt;BD * CE * AF = +DC * EA * FB&lt;/math&gt;&lt;/center&gt;&lt;br&gt;<br /> where all segments in the formula are directed segments.<br /> <br /> == Proof ==<br /> Let &lt;math&gt;{X,Y,Z}&lt;/math&gt; be points on &lt;math&gt;{BC}, {CA}, {AB}&lt;/math&gt; respectively such that &lt;math&gt;AX,BY,CZ&lt;/math&gt; are concurrent, and let &lt;math&gt;{P}&lt;/math&gt; be the point where &lt;math&gt;AX&lt;/math&gt;, &lt;math&gt;BY&lt;/math&gt; and &lt;math&gt;CZ&lt;/math&gt; meet. Draw a parallel to &lt;math&gt;AB&lt;/math&gt; through the point &lt;math&gt;{C}&lt;/math&gt;. Extend &lt;math&gt;AX&lt;/math&gt; until it intersects the parallel at a point &lt;math&gt;\displaystyle{A'}&lt;/math&gt;. Construct &lt;math&gt;\displaystyle{B'}&lt;/math&gt; in a similar way extending &lt;math&gt;BY&lt;/math&gt;.<br /> &lt;center&gt;''(ceva1.png)''&lt;/center&gt;<br /> The triangles &lt;math&gt;{\triangle{ABX}}&lt;/math&gt; and &lt;math&gt;{\triangle{A'CX}}&lt;/math&gt; are similar, and so are &lt;math&gt;\triangle{ABY}&lt;/math&gt; and &lt;math&gt;\triangle{CB'Y}&lt;/math&gt;. Then the following equalities hold:<br /> &lt;math&gt;\begin{displaymath}\frac{BX}{XC}=\frac{AB}{CA'},\qquad\frac{CY}{YA}=\frac{CB'}{BA}\end{displaymath}&lt;/math&gt;<br /> <br /> and thus<br /> &lt;math&gt;\begin{displaymath} \frac{BX}{XC}\cdot\frac{CY}{YA}=\frac{AB}{CA'}\cdot\frac{CB'}{BA}=\frac{CB'}{A'C}. \end{displaymath} (1)&lt;/math&gt;<br /> <br /> Notice that if directed segments are being used then &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;BA&lt;/math&gt; have opposite signs, and therefore when cancelled change the sign of the expression. That's why we changed &lt;math&gt;CA'&lt;/math&gt; to &lt;math&gt;A'C&lt;/math&gt;.<br /> <br /> Now we turn to consider the following similarities: &lt;math&gt;\triangle{AZP}\sim\triangle{A'CP}&lt;/math&gt; and &lt;math&gt;\triangle BZP\sim\triangle B'CP&lt;/math&gt;. From them we get the equalities<br /> &lt;math&gt;\begin{displaymath}\frac{CP}{ZP}=\frac{A'C}{AZ},\qquad\frac{CP}{ZP}=\frac{CB'}{ZB}\end{displaymath}&lt;/math&gt;<br /> <br /> which lead to<br /> &lt;math&gt;\begin{displaymath}\frac{AZ}{ZB}=\frac{A'C}{CB'}.\end{displaymath}&lt;/math&gt;<br /> <br /> Multiplying the last expression with (1) gives<br /> &lt;math&gt;\begin{displaymath}\frac{AZ}{ZB}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1\end{displaymath}&lt;/math&gt;<br /> <br /> and we conclude the proof.<br /> <br /> To prove the converse, suppose that &lt;math&gt;X,Y,Z&lt;/math&gt; are points on &lt;math&gt;{BC, CA, AB}&lt;/math&gt; respectively and satisfying<br /> &lt;math&gt;\begin{displaymath}\frac{AZ}{ZB}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1.\end{displaymath}&lt;/math&gt;<br /> <br /> Let &lt;math&gt;Q&lt;/math&gt; be the intersection point of &lt;math&gt;AX&lt;/math&gt; with &lt;math&gt;BY&lt;/math&gt;, and let &lt;math&gt;Z'&lt;/math&gt; be the intersection of &lt;math&gt;CQ&lt;/math&gt; with &lt;math&gt;AB&lt;/math&gt;. Since then &lt;math&gt;AX,BY,CZ'&lt;/math&gt; are concurrent, we have<br /> &lt;math&gt;\begin{displaymath}\frac{AZ'}{Z'B}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1\end{displaymath}&lt;/math&gt;<br /> <br /> and thus<br /> &lt;math&gt;\begin{displaymath}\frac{AZ'}{Z'B}=\frac{AZ}{ZB}\end{displaymath}&lt;/math&gt;<br /> <br /> which implies &lt;math&gt;Z=Z'&lt;/math&gt;, and therefore &lt;math&gt;AX,BY,CZ&lt;/math&gt; are concurrent.<br /> <br /> == Example ==<br /> Suppose AB, AC, and BC have lengths 13, 14, and 15. If &lt;math&gt;\frac{AF}{FB} = \frac{2}{5}&lt;/math&gt; and &lt;math&gt;\frac{CE}{EA} = \frac{5}{8}&lt;/math&gt;. Find BD and DC.&lt;br&gt;<br /> &lt;br&gt;<br /> If &lt;math&gt;BD = x&lt;/math&gt; and &lt;math&gt;DC = y&lt;/math&gt;, then &lt;math&gt;10x = 40y&lt;/math&gt;, and &lt;math&gt;{x + y = 15}&lt;/math&gt;. From this, we find &lt;math&gt;x = 12&lt;/math&gt; and &lt;math&gt;y = 3&lt;/math&gt;.<br /> <br /> == See also ==<br /> * [[Menelaus' Theorem]]<br /> * [[Stewart's Theorem]]</div> SnowStorm https://artofproblemsolving.com/wiki/index.php?title=Ceva%27s_Theorem&diff=3484 Ceva's Theorem 2006-06-20T20:36:52Z <p>SnowStorm: </p> <hr /> <div>'''Ceva's Theorem''' is an algebraic statement regarding the lengths of [[Cevian|cevians]] in a [[triangle]].<br /> <br /> <br /> == Statement ==<br /> ''(awaiting image)''&lt;br&gt;<br /> A necessary and sufficient condition for AD, BE, CF, where D, E, and F are points of the respective side lines BC, CA, AB of a triangle ABC, to be concurrent is that<br /> &lt;br&gt;&lt;center&gt;&lt;math&gt;BD * CE * AF = +DC * EA * FB&lt;/math&gt;&lt;/center&gt;&lt;br&gt;<br /> where all segments in the formula are directed segments.<br /> <br /> == Proof ==<br /> Let &lt;math&gt;{X,Y,Z}&lt;/math&gt; be points on &lt;math&gt;{BC}, {CA}, {AB}&lt;/math&gt; respectively such that &lt;math&gt;AX,BY,CZ&lt;/math&gt; are concurrent, and let &lt;math&gt;{P}&lt;/math&gt; be the point where &lt;math&gt;AX&lt;/math&gt;, &lt;math&gt;BY&lt;/math&gt; and &lt;math&gt;CZ&lt;/math&gt; meet. Draw a parallel to &lt;math&gt;AB&lt;/math&gt; through the point &lt;math&gt;{C}&lt;/math&gt;. Extend &lt;math&gt;AX&lt;/math&gt; until it intersects the parallel at a point &lt;math&gt;{A'}&lt;/math&gt;. Construct &lt;math&gt;\displaystyle{B'}&lt;/math&gt; in a similar way extending &lt;math&gt;BY&lt;/math&gt;.<br /> &lt;center&gt;''(ceva1.png)''&lt;/center&gt;<br /> The triangles &lt;math&gt;\triangle{ABX}&lt;math&gt; and &lt;math&gt;\triangle{A'CX}&lt;/math&gt; are similar, and so are &lt;math&gt;\triangle{ABY}&lt;/math&gt; and &lt;math&gt;\triangle{CB'Y}&lt;/math&gt;. Then the following equalities hold:<br /> &lt;math&gt;\begin{displaymath}\frac{BX}{XC}=\frac{AB}{CA'},\qquad\frac{CY}{YA}=\frac{CB'}{BA}\end{displaymath}&lt;/math&gt;<br /> <br /> and thus<br /> &lt;math&gt;\begin{displaymath} \frac{BX}{XC}\cdot\frac{CY}{YA}=\frac{AB}{CA'}\cdot\frac{CB'}{BA}=\frac{CB'}{A'C}. \end{displaymath} (1)&lt;/math&gt;<br /> <br /> Notice that if directed segments are being used then &lt;math&gt;AB&lt;/math&gt; and &lt;math&gt;BA&lt;/math&gt; have opposite signs, and therefore when cancelled change the sign of the expression. That's why we changed &lt;math&gt;CA'&lt;/math&gt; to &lt;math&gt;A'C&lt;/math&gt;.<br /> <br /> Now we turn to consider the following similarities: &lt;math&gt;\triangle{AZP}\sim\triangle{A'CP}&lt;/math&gt; and &lt;math&gt;\triangle BZP\sim\triangle B'CP&lt;/math&gt;. From them we get the equalities<br /> &lt;math&gt;\begin{displaymath}\frac{CP}{ZP}=\frac{A'C}{AZ},\qquad\frac{CP}{ZP}=\frac{CB'}{ZB}\end{displaymath}&lt;/math&gt;<br /> <br /> which lead to<br /> &lt;math&gt;\begin{displaymath}\frac{AZ}{ZB}=\frac{A'C}{CB'}.\end{displaymath}&lt;/math&gt;<br /> <br /> Multiplying the last expression with (1) gives<br /> &lt;math&gt;\begin{displaymath}\frac{AZ}{ZB}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1\end{displaymath}&lt;/math&gt;<br /> <br /> and we conclude the proof.<br /> <br /> To prove the converse, suppose that &lt;math&gt;X,Y,Z&lt;/math&gt; are points on &lt;math&gt;{BC, CA, AB}&lt;/math&gt; respectively and satisfying<br /> &lt;math&gt;\begin{displaymath}\frac{AZ}{ZB}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1.\end{displaymath}&lt;/math&gt;<br /> <br /> Let &lt;math&gt;Q&lt;/math&gt; be the intersection point of &lt;math&gt;AX&lt;/math&gt; with &lt;math&gt;BY&lt;/math&gt;, and let &lt;math&gt;Z'&lt;/math&gt; be the intersection of &lt;math&gt;CQ&lt;/math&gt; with &lt;math&gt;AB&lt;/math&gt;. Since then &lt;math&gt;AX,BY,CZ'&lt;/math&gt; are concurrent, we have<br /> &lt;math&gt;\begin{displaymath}\frac{AZ'}{Z'B}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1\end{displaymath}&lt;/math&gt;<br /> <br /> and thus<br /> &lt;math&gt;\begin{displaymath}\frac{AZ'}{Z'B}=\frac{AZ}{ZB}\end{displaymath}&lt;/math&gt;<br /> <br /> which implies &lt;math&gt;Z=Z'&lt;/math&gt;, and therefore &lt;math&gt;AX,BY,CZ&lt;/math&gt; are concurrent.<br /> <br /> == Example ==<br /> Suppose AB, AC, and BC have lengths 13, 14, and 15. If &lt;math&gt;\frac{AF}{FB} = \frac{2}{5}&lt;/math&gt; and &lt;math&gt;\frac{CE}{EA} = \frac{5}{8}&lt;/math&gt;. Find BD and DC.&lt;br&gt;<br /> &lt;br&gt;<br /> If &lt;math&gt;BD = x&lt;/math&gt; and &lt;math&gt;DC = y&lt;/math&gt;, then &lt;math&gt;10x = 40y&lt;/math&gt;, and &lt;math&gt;{x + y = 15}&lt;/math&gt;. From this, we find &lt;math&gt;x = 12&lt;/math&gt; and &lt;math&gt;y = 3&lt;/math&gt;.<br /> <br /> == See also ==<br /> * [[Menelaus' Theorem]]<br /> * [[Stewart's Theorem]]</div> SnowStorm https://artofproblemsolving.com/wiki/index.php?title=User_talk:SnowStorm&diff=3482 User talk:SnowStorm 2006-06-20T20:23:47Z <p>SnowStorm: </p> <hr /> <div>Per Mr. Crawford's suggestion, I will keep a list of my contributions.<br /> &lt;br&gt;&lt;br&gt;<br /> 1) [[Wilson's Theorem]]&lt;br&gt;<br /> 2) Sample of [[Fermat's Little Theorem]]&lt;br&gt;<br /> 3) [[Divisibility]]&lt;br&gt;<br /> 4) [[Ceva's Theorem]]&lt;br&gt;<br /> 5) [[Menelaus' Theorem]]&lt;br&gt;<br /> 6) [[Stewart's Theorem]]&lt;br&gt;<br /> 7) [[Overcounting]]&lt;br&gt;<br /> 8) [[Cevian]]&lt;br&gt;<br /> 9) [[Median]]&lt;br&gt;<br /> 10) [[Altitude]]&lt;br&gt;<br /> 11) [[Angle Bisector]]&lt;br&gt;<br /> <br /> TODO&lt;br&gt;<br /> 4) Add proof, &lt;strike&gt;clean up example&lt;/strike&gt;&lt;br&gt;<br /> 5) Add proof, add example&lt;br&gt;<br /> 6) &lt;strike&gt;Add proof&lt;/strike&gt;, add example&lt;br&gt;<br /> 7) &lt;strike&gt;Find better example&lt;/strike&gt;</div> SnowStorm https://artofproblemsolving.com/wiki/index.php?title=User_talk:SnowStorm&diff=3481 User talk:SnowStorm 2006-06-20T20:23:24Z <p>SnowStorm: </p> <hr /> <div>Per Mr. Crawford's suggestion, I will keep a list of my contributions.<br /> &lt;br&gt;&lt;br&gt;<br /> 1) [[Wilson's Theorem]]&lt;br&gt;<br /> 2) Sample of [[Fermat's Little Theorem]]&lt;br&gt;<br /> 3) [[Divisibility]]&lt;br&gt;<br /> 4) [[Ceva's Theorem]]&lt;br&gt;<br /> 5) [[Menelaus' Theorem]]&lt;br&gt;<br /> 6) [[Stewart's Theorem]]&lt;br&gt;<br /> 7) [[Overcounting]]&lt;br&gt;<br /> 8) [[Cevian]]&lt;br&gt;<br /> 9) [[Median]]&lt;br&gt;<br /> 10) [[Altitude]]&lt;br&gt;<br /> 11) [[Angle Bisector]]&lt;br&gt;<br /> <br /> TODO&lt;br&gt;<br /> 4) Add proof, clean up example&lt;br&gt;<br /> 5) Add proof, add example&lt;br&gt;<br /> 6) &lt;strike&gt;Add proof&lt;/strike&gt;, add example&lt;br&gt;<br /> 7) &lt;strike&gt;Find better example&lt;/strike&gt;</div> SnowStorm https://artofproblemsolving.com/wiki/index.php?title=Cevian&diff=3480 Cevian 2006-06-20T20:20:59Z <p>SnowStorm: </p> <hr /> <div>A cevian is a line segment that extends from one vertex of a [[triangle]] to the opposite side. [[Median|Medians]], [[Altitude|altitudes]], and [[Angle Bisector|angle bisectors]] are all examples of cevians.</div> SnowStorm https://artofproblemsolving.com/wiki/index.php?title=Altitude&diff=3479 Altitude 2006-06-20T20:20:20Z <p>SnowStorm: </p> <hr /> <div>An altitude is a [[cevian]] that is perpendicular to the line segment to which it extends.</div> SnowStorm https://artofproblemsolving.com/wiki/index.php?title=Median_of_a_triangle&diff=3477 Median of a triangle 2006-06-20T20:16:26Z <p>SnowStorm: </p> <hr /> <div>A median is a [[cevian]] which divides one side of a [[polygon]] into two segments of equal length.</div> SnowStorm https://artofproblemsolving.com/wiki/index.php?title=Median_of_a_triangle&diff=3476 Median of a triangle 2006-06-20T20:16:13Z <p>SnowStorm: </p> <hr /> <div>A median is a [[cevian]] which divides one side of a [[polygon]] into two pieces of equal length.</div> SnowStorm https://artofproblemsolving.com/wiki/index.php?title=Cevian&diff=3475 Cevian 2006-06-20T20:14:06Z <p>SnowStorm: </p> <hr /> <div>A cevian is a line segment that extends from one vertex of a [[triangle]] to the opposite side. [[Medians]], [[altitudes]], and [[angle bisectors]] are all examples of cevians.</div> SnowStorm https://artofproblemsolving.com/wiki/index.php?title=Ceva%27s_Theorem&diff=3474 Ceva's Theorem 2006-06-20T20:13:34Z <p>SnowStorm: </p> <hr /> <div>'''Ceva's Theorem''' is an algebraic statement regarding the lengths of [[Cevian|cevians]] in a [[triangle]].<br /> <br /> <br /> == Statement ==<br /> ''(awaiting image)''&lt;br&gt;<br /> A necessary and sufficient condition for AD, BE, CF, where D, E, and F are points of the respective side lines BC, CA, AB of a triangle ABC, to be concurrent is that<br /> &lt;br&gt;&lt;center&gt;&lt;math&gt;BD * CE * AF = +DC * EA * FB&lt;/math&gt;&lt;/center&gt;&lt;br&gt;<br /> where all segments in the formula are directed segments.<br /> <br /> == Example ==<br /> Suppose AB, AC, and BC have lengths 13, 14, and 15. If &lt;math&gt;\frac{AF}{FB} = \frac{2}{5}&lt;/math&gt; and &lt;math&gt;\frac{CE}{EA} = \frac{5}{8}&lt;/math&gt;. Find BD and DC.&lt;br&gt;<br /> &lt;br&gt;<br /> If &lt;math&gt;BD = x&lt;/math&gt; and &lt;math&gt;DC = y&lt;/math&gt;, then &lt;math&gt;10x = 40y&lt;/math&gt;, and &lt;math&gt;{x + y = 15}&lt;/math&gt;. From this, we find &lt;math&gt;x = 12&lt;/math&gt; and &lt;math&gt;y = 3&lt;/math&gt;.<br /> <br /> == See also ==<br /> * [[Menelaus' Theorem]]<br /> * [[Stewart's Theorem]]</div> SnowStorm https://artofproblemsolving.com/wiki/index.php?title=Cevian&diff=3473 Cevian 2006-06-20T20:12:56Z <p>SnowStorm: </p> <hr /> <div>A cevian is a line segment that extends from one vertex of a triangle to the opposite side. [[Medians]], [[altitudes]], and [[angle bisectors]] are all examples of cevians.</div> SnowStorm https://artofproblemsolving.com/wiki/index.php?title=Ceva%27s_Theorem&diff=3472 Ceva's Theorem 2006-06-20T20:10:35Z <p>SnowStorm: /* Example */</p> <hr /> <div>'''Ceva's Theorem''' is an algebraic statement regarding the lengths of [[cevians]] in a [[triangle]].<br /> <br /> <br /> == Statement ==<br /> ''(awaiting image)''&lt;br&gt;<br /> A necessary and sufficient condition for AD, BE, CF, where D, E, and F are points of the respective side lines BC, CA, AB of a triangle ABC, to be concurrent is that<br /> &lt;br&gt;&lt;center&gt;&lt;math&gt;BD * CE * AF = +DC * EA * FB&lt;/math&gt;&lt;/center&gt;&lt;br&gt;<br /> where all segments in the formula are directed segments.<br /> <br /> == Example ==<br /> Suppose AB, AC, and BC have lengths 13, 14, and 15. If &lt;math&gt;\frac{AF}{FB} = \frac{2}{5}&lt;/math&gt; and &lt;math&gt;\frac{CE}{EA} = \frac{5}{8}&lt;/math&gt;. Find BD and DC.&lt;br&gt;<br /> &lt;br&gt;<br /> If &lt;math&gt;BD = x&lt;/math&gt; and &lt;math&gt;DC = y&lt;/math&gt;, then &lt;math&gt;10x = 40y&lt;/math&gt;, and &lt;math&gt;{x + y = 15}&lt;/math&gt;. From this, we find &lt;math&gt;x = 12&lt;/math&gt; and &lt;math&gt;y = 3&lt;/math&gt;.<br /> <br /> == See also ==<br /> * [[Menelaus' Theorem]]<br /> * [[Stewart's Theorem]]</div> SnowStorm https://artofproblemsolving.com/wiki/index.php?title=User_talk:SnowStorm&diff=2964 User talk:SnowStorm 2006-06-19T03:38:55Z <p>SnowStorm: </p> <hr /> <div>Per Mr. Crawford's suggestion, I will keep a list of my contributions.<br /> &lt;br&gt;&lt;br&gt;<br /> 1) [[Wilson's Theorem]]&lt;br&gt;<br /> 2) Sample of [[Fermat's Little Theorem]]&lt;br&gt;<br /> 3) [[Divisibility]]&lt;br&gt;<br /> 4) [[Ceva's Theorem]]&lt;br&gt;<br /> 5) [[Menelaus' Theorem]]&lt;br&gt;<br /> 6) [[Stewart's Theorem]]&lt;br&gt;<br /> 7) [[Overcounting]]&lt;br&gt;<br /> <br /> TODO&lt;br&gt;<br /> 4) Add proof, clean up example&lt;br&gt;<br /> 5) Add proof, add example&lt;br&gt;<br /> 6) Add proof, add example&lt;br&gt;<br /> 7) Find better example</div> SnowStorm https://artofproblemsolving.com/wiki/index.php?title=Overcounting&diff=2963 Overcounting 2006-06-19T03:37:36Z <p>SnowStorm: /* Example */</p> <hr /> <div>== Description ==<br /> Overcounting is the process of counting more than what you need and then systematically subtracting the parts which do not belong.<br /> <br /> == Example ==<br /> Let S be the set of all rational numbers &lt;math&gt;{r}&lt;/math&gt;, &lt;math&gt;0 &lt; r &lt; 1&lt;/math&gt;, that have a repeating decimal expansion in the form &lt;math&gt;0.abcabcabc\dots&lt;/math&gt;, where the digits a, b, and c are not necessarily distinct. To write the elements of S as fractions in lowest terms, how many numerators are required? ([[AIME]] 1992 #5)<br /> <br /> All repeating fractions can be expressed as a number over 999. Hence, there are &lt;math&gt;10*10*10 - 1 = 999&lt;/math&gt; possibilities (since we cannot have zero). We know, however, that some of these numerators have common factors with 999 and will, therefore, be reduced, so we must subtract them from our original count. The prime factorization of 999 is $3^3*37$, so we need to subtract all multiples of 3 and 37;<br /> &lt;br&gt;&lt;center&gt;&lt;math&gt;999 - 333 - 27 = 639&lt;/math&gt;&lt;/center&gt;&lt;br&gt;<br /> But, we have subtracted the multiples of &lt;math&gt;37*3&lt;/math&gt; twice so we must add them back and also we have subtracted the multiples of &lt;math&gt;3^4&lt;/math&gt; which will produce unique numerators so we need to add them back as well. Our final number is thus,<br /> &lt;br&gt;&lt;center&gt;&lt;math&gt;639 + 9 + 12 = 660&lt;/math&gt;&lt;/center&gt;&lt;br&gt;<br /> <br /> ''(This seems to be more inclusion exclusion but it was the best example I could find)''</div> SnowStorm https://artofproblemsolving.com/wiki/index.php?title=Overcounting&diff=2951 Overcounting 2006-06-19T02:47:09Z <p>SnowStorm: </p> <hr /> <div>== Description ==<br /> Overcounting is the process of counting more than what you need and then systematically subtracting the parts which do not belong.<br /> <br /> == Example ==<br /> Let S be the set of all rational numbers &lt;math&gt;{r}&lt;/math&gt;, &lt;math&gt;0 &lt; r &lt; 1&lt;/math&gt;, that have a repeating decimal expansion in the form &lt;math&gt;0.abcabcabc\dots&lt;/math&gt;, where the digits a, b, and c are not necessarily distinct. To write the elements of S as fractions in lowest terms, how many numerators are required?<br /> <br /> ''(Solution Coming)''</div> SnowStorm https://artofproblemsolving.com/wiki/index.php?title=User_talk:SnowStorm&diff=2923 User talk:SnowStorm 2006-06-19T00:54:15Z <p>SnowStorm: </p> <hr /> <div>Per Mr. Crawford's suggestion, I will keep a list of my contributions.<br /> &lt;br&gt;&lt;br&gt;<br /> 1) [[Wilson's Theorem]]&lt;br&gt;<br /> 2) Sample of [[Fermat's Little Theorem]]&lt;br&gt;<br /> 3) [[Divisibility]]&lt;br&gt;<br /> 4) [[Ceva's Theorem]]&lt;br&gt;<br /> 5) [[Menelaus' Theorem]]&lt;br&gt;<br /> 6) [[Stewart's Theorem]]&lt;br&gt;<br /> <br /> TODO&lt;br&gt;<br /> 4) Add proof, clean up example&lt;br&gt;<br /> 5) Add proof, add example&lt;br&gt;<br /> 6) Add proof, add example&lt;br&gt;</div> SnowStorm https://artofproblemsolving.com/wiki/index.php?title=User_talk:SnowStorm&diff=2922 User talk:SnowStorm 2006-06-19T00:53:57Z <p>SnowStorm: </p> <hr /> <div>Per Mr. Crawford's suggestion, I will keep a list of my contributions.<br /> &lt;br&gt;&lt;br&gt;<br /> 1) [[Wilson's Theorem]]&lt;br&gt;<br /> 2) Sample of [[Fermat's Little Theorem]]&lt;br&gt;<br /> 3) [[Divisibility]]&lt;br&gt;<br /> 4) [[Ceva's Theorem]]&lt;br&gt;<br /> 5) [[Menelaus' Theorem]]&lt;br&gt;<br /> 6) [[Stewart's Theorem]]&lt;br&gt;<br /> <br /> TODO<br /> 4) Add proof, clean up example<br /> 5) Add proof, add example<br /> 6) Add proof, add example</div> SnowStorm https://artofproblemsolving.com/wiki/index.php?title=Stewart%27s_Theorem&diff=2921 Stewart's Theorem 2006-06-19T00:52:56Z <p>SnowStorm: </p> <hr /> <div>== Statement ==<br /> ''(awaiting image)''&lt;br&gt;<br /> If a [[cevian]] of length d is drawn and divides side c into segments m and n, then<br /> &lt;br&gt;&lt;center&gt;&lt;math&gt;a^{2}n + b^{2}m = c(d^{2} + mn)&lt;/math&gt;&lt;/center&gt;&lt;br&gt;<br /> <br /> == Proof ==<br /> ''(awaiting addition)''<br /> <br /> == Example ==<br /> ''(awaiting addition)''<br /> <br /> == See also == <br /> * [[Menelaus' Theorem]]<br /> * [[Ceva's Theorem]]</div> SnowStorm https://artofproblemsolving.com/wiki/index.php?title=Ceva%27s_Theorem&diff=2918 Ceva's Theorem 2006-06-19T00:46:04Z <p>SnowStorm: </p> <hr /> <div>'''Ceva's Theorem''' is an algebraic statement regarding the lengths of [[cevians]] in a [[triangle]].<br /> <br /> <br /> == Statement ==<br /> ''(awaiting image)''&lt;br&gt;<br /> A necessary and sufficient condition for AD, BE, CF, where D, E, and F are points of the respective side lines BC, CA, AB of a triangle ABC, to be concurrent is that<br /> &lt;br&gt;&lt;center&gt;&lt;math&gt;BD * CE * AF = +DC * EA * FB&lt;/math&gt;&lt;/center&gt;&lt;br&gt;<br /> where all segments in the formula are directed segments.<br /> <br /> == Example ==<br /> Suppose AB, AC, and BC have lengths 13, 14, and 15. If AF:FB = 2:5 and CE:EA = 5:8. If BD = x and DC = y, then 10x = 40y, and x + y = 15. From this, we find x = 12 and y = 3.<br /> <br /> == See also ==<br /> * [[Menelaus' Theorem]]<br /> * [[Stewart's Theorem]]</div> SnowStorm https://artofproblemsolving.com/wiki/index.php?title=Menelaus%27_Theorem&diff=2916 Menelaus' Theorem 2006-06-19T00:44:43Z <p>SnowStorm: </p> <hr /> <div>== Statement ==<br /> ''(awaiting image)''<br /> A necessary and sufficient condition for points D, E, F on the respective side lines BC, CA, AB of a triangle ABC to be collinear is that<br /> &lt;br&gt;&lt;center&gt;&lt;math&gt;BD*CE*AF = -DC*EA*FB&lt;/math&gt;&lt;/center&gt;&lt;br&gt;<br /> where all segments in the formula are directed segments.<br /> <br /> <br /> == Example ==<br /> ''Does anyone have a good example? ''<br /> <br /> <br /> == See also ==<br /> * [[Ceva's Theorem]]<br /> * [[Stewart's Theorem]]</div> SnowStorm https://artofproblemsolving.com/wiki/index.php?title=User_talk:SnowStorm&diff=2913 User talk:SnowStorm 2006-06-19T00:42:06Z <p>SnowStorm: </p> <hr /> <div>Per Mr. Crawford's suggestion, I will keep a list of my contributions.<br /> &lt;br&gt;&lt;br&gt;<br /> 1) [[Wilson's Theorem]]&lt;br&gt;<br /> 2) Sample of [[Fermat's Little Theorem]]&lt;br&gt;<br /> 3) [[Divisibility]]&lt;br&gt;<br /> 4) [[Ceva's Theorem]]&lt;br&gt;<br /> 5) [[Menelaus' Theorem]]&lt;br&gt;</div> SnowStorm https://artofproblemsolving.com/wiki/index.php?title=Menelaus%27_Theorem&diff=2912 Menelaus' Theorem 2006-06-19T00:41:30Z <p>SnowStorm: </p> <hr /> <div>== Statement ==<br /> ''(awaiting image)''<br /> A necessary and sufficient condition for points D, E, F on the respective side lines BC, CA, AB of a triangle ABC to be collinear is that<br /> &lt;br&gt;&lt;center&gt;&lt;math&gt;BD*CE*AF = -DC*EA*FB&lt;/math&gt;&lt;/center&gt;&lt;br&gt;<br /> where all segments in the formula are directed segments.<br /> <br /> <br /> == Example ==<br /> ''Does anyone have a good example? ''</div> SnowStorm https://artofproblemsolving.com/wiki/index.php?title=Ceva%27s_Theorem&diff=2906 Ceva's Theorem 2006-06-19T00:35:47Z <p>SnowStorm: /* Statement */</p> <hr /> <div>== Statement ==<br /> ''(awaiting image)''&lt;br&gt;<br /> A necessary and sufficient condition for AD, BE, CF, where D, E, and F are points of the respective side lines BC, CA, AB of a triangle ABC, to be concurrent is that<br /> &lt;br&gt;&lt;center&gt;&lt;math&gt;BD * CE * AF = +DC * EA * FB&lt;/math&gt;&lt;/center&gt;&lt;br&gt;<br /> where all segments in the formula are directed segments.<br /> <br /> == Example ==<br /> Suppose AB, AC, and BC have lengths 13, 14, and 15. If AF:FB = 2:5 and CE:EA = 5:8. If BD = x and DC = y, then 10x = 40y, and x + y = 15. From this, we find x = 12 and y = 3.</div> SnowStorm https://artofproblemsolving.com/wiki/index.php?title=Ceva%27s_Theorem&diff=2904 Ceva's Theorem 2006-06-19T00:35:35Z <p>SnowStorm: /* Statement */</p> <hr /> <div>== Statement ==<br /> ''(awaiting image)''<br /> A necessary and sufficient condition for AD, BE, CF, where D, E, and F are points of the respective side lines BC, CA, AB of a triangle ABC, to be concurrent is that<br /> &lt;br&gt;&lt;center&gt;&lt;math&gt;BD * CE * AF = +DC * EA * FB&lt;/math&gt;&lt;/center&gt;&lt;br&gt;<br /> where all segments in the formula are directed segments.<br /> <br /> == Example ==<br /> Suppose AB, AC, and BC have lengths 13, 14, and 15. If AF:FB = 2:5 and CE:EA = 5:8. If BD = x and DC = y, then 10x = 40y, and x + y = 15. From this, we find x = 12 and y = 3.</div> SnowStorm https://artofproblemsolving.com/wiki/index.php?title=Ceva%27s_Theorem&diff=2903 Ceva's Theorem 2006-06-19T00:35:14Z <p>SnowStorm: /* Statement */</p> <hr /> <div>== Statement ==<br /> ***awaiting image***<br /> A necessary and sufficient condition for AD, BE, CF, where D, E, and F are points of the respective side lines BC, CA, AB of a triangle ABC, to be concurrent is that<br /> &lt;br&gt;&lt;center&gt;&lt;math&gt;BD * CE * AF = +DC * EA * FB&lt;/math&gt;&lt;/center&gt;&lt;br&gt;<br /> where all segments in the formula are directed segments.<br /> <br /> == Example ==<br /> Suppose AB, AC, and BC have lengths 13, 14, and 15. If AF:FB = 2:5 and CE:EA = 5:8. If BD = x and DC = y, then 10x = 40y, and x + y = 15. From this, we find x = 12 and y = 3.</div> SnowStorm https://artofproblemsolving.com/wiki/index.php?title=User_talk:SnowStorm&diff=2902 User talk:SnowStorm 2006-06-19T00:34:41Z <p>SnowStorm: </p> <hr /> <div>Per Mr. Crawford's suggestion, I will keep a list of my contributions.<br /> &lt;br&gt;&lt;br&gt;<br /> 1) [[Wilson's Theorem]]&lt;br&gt;<br /> 2) Sample of [[Fermat's Little Theorem]]&lt;br&gt;<br /> 3) [[Divisibility]]&lt;br&gt;<br /> 4) [[Ceva's Theorem]]&lt;br&gt;</div> SnowStorm https://artofproblemsolving.com/wiki/index.php?title=Ceva%27s_Theorem&diff=2809 Ceva's Theorem 2006-06-18T21:24:44Z <p>SnowStorm: </p> <hr /> <div>== Statement ==<br /> <br /> A necessary and sufficient condition for AD, BE, CF, where D, E, and F are points of the respective side lines BC, CA, AB of a triangle ABC, to be concurrent is that<br /> &lt;br&gt;&lt;center&gt;&lt;math&gt;BD * CE * AF = +DC * EA * FB&lt;/math&gt;&lt;/center&gt;&lt;br&gt;<br /> where all segments in the formula are directed segments.<br /> <br /> == Example ==<br /> Suppose AB, AC, and BC have lengths 13, 14, and 15. If AF:FB = 2:5 and CE:EA = 5:8. If BD = x and DC = y, then 10x = 40y, and x + y = 15. From this, we find x = 12 and y = 3.</div> SnowStorm https://artofproblemsolving.com/wiki/index.php?title=User_talk:SnowStorm&diff=2806 User talk:SnowStorm 2006-06-18T21:15:22Z <p>SnowStorm: </p> <hr /> <div>Per Mr. Crawford's suggestion, I will keep a list of my contributions.<br /> &lt;br&gt;&lt;br&gt;<br /> 1) [[Wilson's Theorem]]&lt;br&gt;<br /> 2) Sample of [[Fermat's Little Theorem]]&lt;br&gt;<br /> 3) [[Divisibility]]&lt;br&gt;</div> SnowStorm https://artofproblemsolving.com/wiki/index.php?title=User_talk:SnowStorm&diff=2805 User talk:SnowStorm 2006-06-18T21:14:41Z <p>SnowStorm: </p> <hr /> <div>Per Mr. Crawford's suggestion, I will keep a list of my contributions.<br /> &lt;br&gt;&lt;br&gt;<br /> 1) [[Wilson's Theorem]]&lt;br&gt;<br /> 2) Sample of [[Fermat's Little Theorem]]&lt;br&gt;<br /> 3) [[Divisibilty]]&lt;br&gt;</div> SnowStorm https://artofproblemsolving.com/wiki/index.php?title=User_talk:SnowStorm&diff=2802 User talk:SnowStorm 2006-06-18T21:04:40Z <p>SnowStorm: </p> <hr /> <div>Per Mr. Crawford's suggestion, I will keep a list of my contributions.<br /> &lt;br&gt;&lt;br&gt;<br /> 1) Wilson's Theorem&lt;br&gt;<br /> 2) Sample of Fermat's Little Theorem&lt;br&gt;<br /> 3) Divisibilty&lt;br&gt;</div> SnowStorm https://artofproblemsolving.com/wiki/index.php?title=Divisibility&diff=2801 Divisibility 2006-06-18T21:03:53Z <p>SnowStorm: </p> <hr /> <div>== Description ==<br /> Divisibility is the ability of a number to be evenly divided by another number. For example, four divided by two is equal to two, and therefore, four is divisible by two.<br /> <br /> == By &lt;math&gt;2^n&lt;/math&gt; ==<br /> A number is divisible by &lt;math&gt;2^n&lt;/math&gt; if the last &lt;math&gt;{n}&lt;/math&gt; digits of the number are divisible by &lt;math&gt;2^n&lt;/math&gt;.<br /> <br /> == By 3 ==<br /> A number is divisible by 3 if the sum of its digits is divisible by 3.<br /> <br /> == By 9 ==<br /> A number is divisible by 9 if the sum of its digits is divisible by 9.<br /> <br /> == By 7 ==<br /> Rule 1: Partition &lt;math&gt;n&lt;/math&gt; into 3 digit numbers from the right (&lt;math&gt;d_3d_2d_1,d_6d_5d_4,\dots&lt;/math&gt;). If the alternating sum (&lt;math&gt;d_3d_2d_1 - d_6d_5d_4 + d_9d_8d_7 - \dots&lt;/math&gt;) is divisible by 7 then the number is divisible by 7.&lt;br&gt;<br /> &lt;br&gt;<br /> Rule 2: Truncate the last digit of &lt;math&gt;{n}&lt;/math&gt;, and subtract twice that digit from the remaining number. If the result is divisible by 7, then the number is divisible by 7. This process can be repeated for large numbers.&lt;br&gt;<br /> <br /> == By 11 ==<br /> A number is divisible by 11 if the alternating sum of the digits is divisible by 11.<br /> <br /> == By 13 ==<br /> See rule 1 for divisibility by 7, a number is divisible by 13 if the same specified sum is divisible by 13.</div> SnowStorm https://artofproblemsolving.com/wiki/index.php?title=User_talk:SnowStorm&diff=2794 User talk:SnowStorm 2006-06-18T20:48:42Z <p>SnowStorm: </p> <hr /> <div>Per Mr. Crawford's suggestion, I will keep a list of my contributions.<br /> &lt;br&gt;&lt;br&gt;<br /> 1) Wilson's Theorem&lt;br&gt;<br /> 2) Sample of Fermat's Little Theorem&lt;br&gt;</div> SnowStorm https://artofproblemsolving.com/wiki/index.php?title=User_talk:SnowStorm&diff=2792 User talk:SnowStorm 2006-06-18T20:46:05Z <p>SnowStorm: </p> <hr /> <div>Per Mr. Crawford's suggestion, I will keep a list of my contributions.<br /> <br /> 1) Wilson's Theorem<br /> 2) Sample of Fermat's Little Theorem</div> SnowStorm https://artofproblemsolving.com/wiki/index.php?title=Fermat%27s_Little_Theorem&diff=2662 Fermat's Little Theorem 2006-06-18T18:14:41Z <p>SnowStorm: /* Example */</p> <hr /> <div>=== Statement ===<br /> <br /> If &lt;math&gt;{a}&lt;/math&gt; is an [[integer]] and &lt;math&gt;{p}&lt;/math&gt; is a [[prime]] number, then &lt;math&gt;a^{p-1}\equiv 1 \pmod {p}&lt;/math&gt;.<br /> <br /> Note: This theorem is a special case of [[Euler's totient theorem]].<br /> <br /> === Corollary ===<br /> <br /> A frequently used corolary of Fermat's little theorem is &lt;math&gt; a^p \equiv a \pmod {p}&lt;/math&gt;.<br /> As you can see, it is derived by multipling both sides of the theorem by a.<br /> <br /> === Sample Problem ===<br /> One of Euler's conjectures was disproved in then 1960s by three American mathematicians when they showed there was a positive integer such that &lt;math&gt;133^5+110^5+84^5+27^5=n^5&lt;/math&gt;. Find the value of &lt;math&gt;{n}&lt;/math&gt;. ([[AIME]] 1989 #9)&lt;br&gt;&lt;br&gt;<br /> <br /> By Fermat's Little Theorem, we know &lt;math&gt;{n^{5}}&lt;/math&gt; is congruent to &lt;math&gt;n&lt;/math&gt; [[modulo]] 5. Hence,&lt;br&gt;<br /> &lt;center&gt;&lt;math&gt;3 + 0 + 4 + 7 \equiv n\pmod{5}&lt;/math&gt;&lt;/center&gt;&lt;br&gt;<br /> &lt;center&gt;&lt;math&gt;4 \equiv n\pmod{5}&lt;/math&gt;&lt;/center&gt;&lt;br&gt;<br /> Continuing, we examine the equation modulo 3,<br /> &lt;center&gt;&lt;math&gt;-1 + 1 + 0 + 0 \equiv n\pmod{3}&lt;/math&gt;&lt;/center&gt;&lt;br&gt;<br /> &lt;center&gt;&lt;math&gt;0 \equiv n\pmod{3}&lt;/math&gt;&lt;/center&gt;&lt;br&gt;<br /> Thus, &lt;math&gt;n&lt;/math&gt; is divisible by three and leaves a remainder of four when divided by 5. It's obvious that &lt;math&gt;n&gt;133&lt;/math&gt; so the only possibilities are &lt;math&gt;n = 144&lt;/math&gt; or &lt;math&gt;n = 174&lt;/math&gt;. It quickly becomes apparent that 174 is much too large so &lt;math&gt;{n}&lt;/math&gt; must be 144.<br /> <br /> === Credit ===<br /> <br /> This theorem is credited to [[Pierre de Fermat]].<br /> <br /> === See also ===<br /> <br /> * [[Number theory]]<br /> * [[Modular arithmetic]]<br /> * [[Euler's phi function]]<br /> * [[Euler's totient theorem]]</div> SnowStorm https://artofproblemsolving.com/wiki/index.php?title=Fermat%27s_Little_Theorem&diff=2661 Fermat's Little Theorem 2006-06-18T18:14:23Z <p>SnowStorm: </p> <hr /> <div>=== Statement ===<br /> <br /> If &lt;math&gt;{a}&lt;/math&gt; is an [[integer]] and &lt;math&gt;{p}&lt;/math&gt; is a [[prime]] number, then &lt;math&gt;a^{p-1}\equiv 1 \pmod {p}&lt;/math&gt;.<br /> <br /> Note: This theorem is a special case of [[Euler's totient theorem]].<br /> <br /> === Corollary ===<br /> <br /> A frequently used corolary of Fermat's little theorem is &lt;math&gt; a^p \equiv a \pmod {p}&lt;/math&gt;.<br /> As you can see, it is derived by multipling both sides of the theorem by a.<br /> <br /> === Example ===<br /> One of Euler's conjectures was disproved in then 1960s by three American mathematicians when they showed there was a positive integer such that &lt;math&gt;133^5+110^5+84^5+27^5=n^5&lt;/math&gt;. Find the value of &lt;math&gt;{n}&lt;/math&gt;. ([[AIME]] 1989 #9)&lt;br&gt;&lt;br&gt;<br /> <br /> By Fermat's Little Theorem, we know &lt;math&gt;{n^{5}}&lt;/math&gt; is congruent to &lt;math&gt;n&lt;/math&gt; [[modulo]] 5. Hence,&lt;br&gt;<br /> &lt;center&gt;&lt;math&gt;3 + 0 + 4 + 7 \equiv n\pmod{5}&lt;/math&gt;&lt;/center&gt;&lt;br&gt;<br /> &lt;center&gt;&lt;math&gt;4 \equiv n\pmod{5}&lt;/math&gt;&lt;/center&gt;&lt;br&gt;<br /> Continuing, we examine the equation modulo 3,<br /> &lt;center&gt;&lt;math&gt;-1 + 1 + 0 + 0 \equiv n\pmod{3}&lt;/math&gt;&lt;/center&gt;&lt;br&gt;<br /> &lt;center&gt;&lt;math&gt;0 \equiv n\pmod{3}&lt;/math&gt;&lt;/center&gt;&lt;br&gt;<br /> Thus, &lt;math&gt;n&lt;/math&gt; is divisible by three and leaves a remainder of four when divided by 5. It's obvious that &lt;math&gt;n&gt;133&lt;/math&gt; so the only possibilities are &lt;math&gt;n = 144&lt;/math&gt; or &lt;math&gt;n = 174&lt;/math&gt;. It quickly becomes apparent that 174 is much too large so &lt;math&gt;{n}&lt;/math&gt; must be 144.<br /> <br /> === Credit ===<br /> <br /> This theorem is credited to [[Pierre de Fermat]].<br /> <br /> === See also ===<br /> <br /> * [[Number theory]]<br /> * [[Modular arithmetic]]<br /> * [[Euler's phi function]]<br /> * [[Euler's totient theorem]]</div> SnowStorm https://artofproblemsolving.com/wiki/index.php?title=Wilson%27s_Theorem&diff=2616 Wilson's Theorem 2006-06-18T15:41:41Z <p>SnowStorm: /* Proof */</p> <hr /> <div>== Statement ==<br /> If and only if &lt;math&gt;{p}&lt;/math&gt; is a [[prime]], then &lt;math&gt;(p-1)! + 1&lt;/math&gt; is a multiple of &lt;math&gt;{p}&lt;/math&gt;. In other words &lt;math&gt;(p-1)! \equiv -1 \pmod{p}&lt;/math&gt;.<br /> <br /> == Proof ==<br /> Wilson's theorem is easily verifiable for 2 and 3, so let's consider &lt;math&gt;p&gt;3&lt;/math&gt;. If &lt;math&gt;{p}&lt;/math&gt; is composite, then its positive factors are among <br /> &lt;center&gt;&lt;math&gt;1, 2, 3, \dots, p-1&lt;/math&gt;.&lt;/center&gt; &lt;br&gt;<br /> Hence, &lt;math&gt;\gcd( (p - 1)!, p) &gt; 1&lt;/math&gt;, so &lt;math&gt;(p-1)! \neq -1 \pmod{p}&lt;/math&gt;.<br /> <br /> However if &lt;math&gt;{p}&lt;/math&gt; is prime, then each of the above integers are relatively [[prime]] to &lt;math&gt;{p}&lt;/math&gt;. So for each of these integers a there is another &lt;math&gt;b&lt;/math&gt; such that &lt;math&gt;ab \equiv 1 \pmod{p}&lt;/math&gt;. It is important to note that this &lt;math&gt;b&lt;/math&gt; is unique [[modulo]] &lt;math&gt;{p}&lt;/math&gt;, and that since &lt;math&gt;{p}&lt;/math&gt; is prime, &lt;math&gt;a = b&lt;/math&gt; if and only if &lt;math&gt;{a}&lt;/math&gt; is &lt;math&gt;1&lt;/math&gt; or &lt;math&gt;p-1&lt;/math&gt;. Now if we omit 1 and &lt;math&gt;p-1&lt;/math&gt;, then the others can be grouped into pairs whose product is congruent to one, <br /> &lt;center&gt;&lt;math&gt;2\cdot3\cdot4\cdots(p-2) \equiv 1\pmod{p}&lt;/math&gt;&lt;/center&gt; &lt;br&gt;<br /> <br /> Finally, multiply this equality by &lt;math&gt;p-1&lt;/math&gt; to complete the proof.<br /> <br /> ==Example==<br /> Let &lt;math&gt;{p}&lt;/math&gt; be a prime number such that dividing &lt;math&gt;{p}&lt;/math&gt; by 4 leaves the remainder 1. Show that there is an integer &lt;math&gt;{n}&lt;/math&gt; such that &lt;math&gt;n^2 + 1&lt;/math&gt; is divisible by &lt;math&gt;{p}&lt;/math&gt;.<br /> <br /> &lt;Solutions?&gt;<br /> <br /> == See also ==<br /> * [[Number theory]]<br /> * [[Modular arithmetic]]<br /> * [[Factorial]]</div> SnowStorm https://artofproblemsolving.com/wiki/index.php?title=Wilson%27s_Theorem&diff=2615 Wilson's Theorem 2006-06-18T15:40:37Z <p>SnowStorm: /* Statement */</p> <hr /> <div>== Statement ==<br /> If and only if &lt;math&gt;{p}&lt;/math&gt; is a [[prime]], then &lt;math&gt;(p-1)! + 1&lt;/math&gt; is a multiple of &lt;math&gt;{p}&lt;/math&gt;. In other words &lt;math&gt;(p-1)! \equiv -1 \pmod{p}&lt;/math&gt;.<br /> <br /> == Proof ==<br /> Wilson's theorem is easily verifiable for 2 and 3, so let's consider &lt;math&gt;p&gt;3&lt;/math&gt;. If &lt;math&gt;{p}&lt;/math&gt; is composite, then its positive factors are among <br /> &lt;center&gt;&lt;math&gt;1, 2, 3, \dots, p-1&lt;/math&gt;.&lt;/center&gt; &lt;br&gt;<br /> Hence, &lt;math&gt;\gcd( (p - 1)!, p) &gt; 1&lt;/math&gt;, so &lt;math&gt;(p-1)! \neq -1 \pmod{p}&lt;/math&gt;.<br /> <br /> However if &lt;math&gt;{p}&lt;/math&gt; is prime, then each of the above integers are relatively prime to &lt;math&gt;{p}&lt;/math&gt;. So for each of these integers a there is another &lt;math&gt;b&lt;/math&gt; such that &lt;math&gt;ab \equiv 1 \pmod{p}&lt;/math&gt;. It is important to note that this &lt;math&gt;b&lt;/math&gt; is unique modulo &lt;math&gt;{p}&lt;/math&gt;, and that since &lt;math&gt;{p}&lt;/math&gt; is prime, &lt;math&gt;a = b&lt;/math&gt; if and only if &lt;math&gt;{a}&lt;/math&gt; is &lt;math&gt;1&lt;/math&gt; or &lt;math&gt;p-1&lt;/math&gt;. Now if we omit 1 and &lt;math&gt;p-1&lt;/math&gt;, then the others can be grouped into pairs whose product is congruent to one, <br /> &lt;center&gt;&lt;math&gt;2\cdot3\cdot4\cdots(p-2) \equiv 1\pmod{p}&lt;/math&gt;&lt;/center&gt; &lt;br&gt;<br /> <br /> Finally, multiply this equality by &lt;math&gt;p-1&lt;/math&gt; to complete the proof.<br /> <br /> ==Example==<br /> Let &lt;math&gt;{p}&lt;/math&gt; be a prime number such that dividing &lt;math&gt;{p}&lt;/math&gt; by 4 leaves the remainder 1. Show that there is an integer &lt;math&gt;{n}&lt;/math&gt; such that &lt;math&gt;n^2 + 1&lt;/math&gt; is divisible by &lt;math&gt;{p}&lt;/math&gt;.<br /> <br /> &lt;Solutions?&gt;<br /> <br /> == See also ==<br /> * [[Number theory]]<br /> * [[Modular arithmetic]]<br /> * [[Factorial]]</div> SnowStorm https://artofproblemsolving.com/wiki/index.php?title=Wilson%27s_Theorem&diff=2614 Wilson's Theorem 2006-06-18T15:39:46Z <p>SnowStorm: /* Proof */</p> <hr /> <div>== Statement ==<br /> If and only if &lt;math&gt;{p}&lt;/math&gt; is a prime, then &lt;math&gt;(p-1)! + 1&lt;/math&gt; is a multiple of &lt;math&gt;{p}&lt;/math&gt;. In other words &lt;math&gt;(p-1)! \equiv -1 \pmod{p}&lt;/math&gt;.<br /> <br /> == Proof ==<br /> Wilson's theorem is easily verifiable for 2 and 3, so let's consider &lt;math&gt;p&gt;3&lt;/math&gt;. If &lt;math&gt;{p}&lt;/math&gt; is composite, then its positive factors are among <br /> &lt;center&gt;&lt;math&gt;1, 2, 3, \dots, p-1&lt;/math&gt;.&lt;/center&gt; &lt;br&gt;<br /> Hence, &lt;math&gt;\gcd( (p - 1)!, p) &gt; 1&lt;/math&gt;, so &lt;math&gt;(p-1)! \neq -1 \pmod{p}&lt;/math&gt;.<br /> <br /> However if &lt;math&gt;{p}&lt;/math&gt; is prime, then each of the above integers are relatively prime to &lt;math&gt;{p}&lt;/math&gt;. So for each of these integers a there is another &lt;math&gt;b&lt;/math&gt; such that &lt;math&gt;ab \equiv 1 \pmod{p}&lt;/math&gt;. It is important to note that this &lt;math&gt;b&lt;/math&gt; is unique modulo &lt;math&gt;{p}&lt;/math&gt;, and that since &lt;math&gt;{p}&lt;/math&gt; is prime, &lt;math&gt;a = b&lt;/math&gt; if and only if &lt;math&gt;{a}&lt;/math&gt; is &lt;math&gt;1&lt;/math&gt; or &lt;math&gt;p-1&lt;/math&gt;. Now if we omit 1 and &lt;math&gt;p-1&lt;/math&gt;, then the others can be grouped into pairs whose product is congruent to one, <br /> &lt;center&gt;&lt;math&gt;2\cdot3\cdot4\cdots(p-2) \equiv 1\pmod{p}&lt;/math&gt;&lt;/center&gt; &lt;br&gt;<br /> <br /> Finally, multiply this equality by &lt;math&gt;p-1&lt;/math&gt; to complete the proof.<br /> <br /> ==Example==<br /> Let &lt;math&gt;{p}&lt;/math&gt; be a prime number such that dividing &lt;math&gt;{p}&lt;/math&gt; by 4 leaves the remainder 1. Show that there is an integer &lt;math&gt;{n}&lt;/math&gt; such that &lt;math&gt;n^2 + 1&lt;/math&gt; is divisible by &lt;math&gt;{p}&lt;/math&gt;.<br /> <br /> &lt;Solutions?&gt;<br /> <br /> == See also ==<br /> * [[Number theory]]<br /> * [[Modular arithmetic]]<br /> * [[Factorial]]</div> SnowStorm https://artofproblemsolving.com/wiki/index.php?title=Wilson%27s_Theorem&diff=2295 Wilson's Theorem 2006-06-17T19:19:30Z <p>SnowStorm: /* Proof */</p> <hr /> <div>== Statement ==<br /> If and only if p is a prime, then &lt;math&gt;(p-1)! + 1&lt;/math&gt; is a multiple of p. Written more mathematically,<br /> &lt;math&gt;(p-1)! \equiv -1 \pmod{p}&lt;/math&gt;<br /> <br /> == Proof ==<br /> Wilson's theorem is easily verifiable for 2 and 3, so let's consider &lt;math&gt;p&gt;3&lt;/math&gt;. If p is composite, then its positive factors are among<br /> &lt;math&gt;1, 2, 3, \dots, p-1&lt;/math&gt;<br /> Hence, &lt;math&gt;gcd((p-1)!, p) &gt; 1&lt;/math&gt;, so &lt;math&gt;(p-1)! \neq -1 \pmod{p}&lt;/math&gt;.<br /> However if &lt;math&gt;p&lt;/math&gt; is prime, then each of the above integers are relatively prime to &lt;math&gt;p&lt;/math&gt;. So for each of these integers &lt;math&gt;a&lt;/math&gt; there is another &lt;math&gt;b&lt;/math&gt; such that &lt;math&gt;ab \equiv 1 \pmod{p}&lt;/math&gt;. It is important to note that this &lt;math&gt;b&lt;/math&gt; is unique modulo &lt;math&gt;p&lt;/math&gt;, and that since &lt;math&gt;p&lt;/math&gt; is prime, &lt;math&gt;a = b&lt;/math&gt; if and only if &lt;math&gt;a&lt;/math&gt; is &lt;math&gt;1&lt;/math&gt; or &lt;math&gt;p-1&lt;/math&gt;. Now if we omit &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;p-1&lt;/math&gt;, then the others can be grouped into pairs whose product is congruent to one,<br /> &lt;math&gt;2*3*4*\dots*(p-2) \equiv 1\pmod{p}&lt;/math&gt;<br /> Finally, multiply this equality by p-1 to complete the proof.<br /> Insert non-formatted text here</div> SnowStorm https://artofproblemsolving.com/wiki/index.php?title=Wilson%27s_Theorem&diff=2293 Wilson's Theorem 2006-06-17T19:15:12Z <p>SnowStorm: </p> <hr /> <div>== Statement ==<br /> If and only if p is a prime, then &lt;math&gt;(p-1)! + 1&lt;/math&gt; is a multiple of p. Written more mathematically,<br /> &lt;math&gt;(p-1)! \equiv -1 \pmod{p}&lt;/math&gt;<br /> <br /> == Proof ==<br /> Wilson's theorem is easily verifiable for 2 and 3, so let's consider &lt;math&gt;p&gt;3&lt;/math&gt;. If &lt;math&gt;p&lt;/math&gt; is composite, then its positive factors are among<br /> &lt;math&gt;1, 2, 3, \dots, p-1&lt;/math&gt;<br /> Hence, &lt;math&gt;gcd((p-1)!,p) &gt; 1&lt;/math&gt;, so &lt;math&gt;(p-1)! \neq -1 \pmod{p}&lt;/math&gt;.<br /> However if &lt;math&gt;p&lt;/math&gt; is prime, then each of the above integers are relatively prime to &lt;math&gt;p&lt;/math&gt;. So for each of these integers &lt;math&gt;a&lt;/math&gt; there is another &lt;math&gt;b&lt;/math&gt; such that &lt;math&gt;ab \equiv 1 \pmod{p}&lt;/math&gt;. It is important to note that this &lt;math&gt;b&lt;/math&gt; is unique modulo &lt;math&gt;p&lt;/math&gt;, and that since &lt;math&gt;p&lt;/math&gt; is prime, &lt;math&gt;a = b&lt;/math&gt; if and only if &lt;math&gt;a&lt;/math&gt; is &lt;math&gt;1&lt;/math&gt; or &lt;math&gt;p-1&lt;/math&gt;. Now if we omit &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;p-1&lt;/math&gt;, then the others can be grouped into pairs whose product is congruent to one,<br /> &lt;math&gt;2*3*4*\dots*(p-2) \equiv 1\pmod{p}&lt;/math&gt;<br /> Finally, multiply this equality by p-1 to complete the proof.<br /> Insert non-formatted text here</div> SnowStorm