tag:blogger.com,1999:blog-6526388641946874429.post5251294664749785116..comments2023-07-06T09:17:00.236-06:00Comments on Olimpiada de Matemáticas en Chihuahua: Problema del MartesDavid (sirio11)http://www.blogger.com/profile/13765612869477578855noreply@blogger.comBlogger5125tag:blogger.com,1999:blog-6526388641946874429.post-13352001963331773372013-10-03T08:11:51.608-06:002013-10-03T08:11:51.608-06:00Los puntos estan en el segmento?Los puntos estan en el segmento?Anonymoushttps://www.blogger.com/profile/04851070136194676975noreply@blogger.comtag:blogger.com,1999:blog-6526388641946874429.post-4805196080459697712013-10-02T22:14:17.075-06:002013-10-02T22:14:17.075-06:00SiSiPepe Floreshttps://www.blogger.com/profile/13137679245007037237noreply@blogger.comtag:blogger.com,1999:blog-6526388641946874429.post-26732202960320930272013-10-02T10:23:19.978-06:002013-10-02T10:23:19.978-06:00Los puntos estan en el segmento?Los puntos estan en el segmento?Anonymoushttps://www.blogger.com/profile/04851070136194676975noreply@blogger.comtag:blogger.com,1999:blog-6526388641946874429.post-13385423329104007452013-10-01T23:46:54.318-06:002013-10-01T23:46:54.318-06:00Bien, Chacón Bien, Chacón Pepe Floreshttps://www.blogger.com/profile/13137679245007037237noreply@blogger.comtag:blogger.com,1999:blog-6526388641946874429.post-58006647593962106792013-10-01T21:53:33.016-06:002013-10-01T21:53:33.016-06:00Llamo a los dos puntos $B,C$ con $B$ más cerca de ...Llamo a los dos puntos $B,C$ con $B$ más cerca de $A$ que $C$ y a la longitud $AD$ la llamo $a$.<br />Se puede formar un triángulo si y sólo si<br />$AB < BC+CD=BD; BC<AB+CD; CD<AB+BC=AC$, esto es equivalente a<br />$AB< \frac{a}{2};$ $CD< \frac{a}{2};$ $\frac{a}{2}<AB+CD$ (esto es porque<br />$AB+BC+CD=a$).<br />A cada pareja de puntos que se escoja en $AD$ le asignamos el Luis Chacónhttps://www.blogger.com/profile/07203499159766102010noreply@blogger.com