tag:blogger.com,1999:blog-42317775926112977532024-03-13T22:46:45.338-07:00PERÚ GEOMÉTRICOUnknownnoreply@blogger.comBlogger22125tag:blogger.com,1999:blog-4231777592611297753.post-11196990259571397272013-12-05T08:03:00.002-08:002013-12-05T08:04:14.800-08:00ON-2 PB15<div class="separator" style="clear: both; text-align: center;">
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<br />Unknownnoreply@blogger.com1tag:blogger.com,1999:blog-4231777592611297753.post-22656303325357267302013-11-04T16:24:00.000-08:002013-11-06T08:40:16.741-08:00Nivel UNI Problema 18: 13 solucionesNivel UNI Problema 18: 13 diferentes soluciones de parte de nuestros miembros y colaboradores
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Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-4231777592611297753.post-28737393023107389372013-10-17T14:58:00.002-07:002013-10-17T15:00:46.081-07:00Problema de Inaguración Sección Universitario NU-P0<div class="separator" style="clear: both; text-align: center;">
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<a href="http://perugeometricouniversiratio.blogspot.com/2013/10/nu-p0.html" target="_blank">Enlace relacionado</a>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-4231777592611297753.post-22217047644943203672013-10-17T14:57:00.000-07:002013-10-17T14:57:06.674-07:00Problema de la Sección Nivel universitario NU-P1<div class="separator" style="clear: both; text-align: center;">
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<a href="http://perugeometricouniversiratio.blogspot.com/2013/10/nu-p1.html" target="_blank">Enlace relacionado</a>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-4231777592611297753.post-72147950852730931302013-10-17T14:55:00.000-07:002013-10-17T14:55:02.407-07:00Sección Nivel Universitario<div class="separator" style="clear: both; text-align: center;">
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<a href="http://perugeometricouniversiratio.blogspot.com/2013/10/nu-p2.html" target="_blank">Enlace relacionado</a>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-4231777592611297753.post-40389180658876429142013-07-06T11:42:00.000-07:002013-07-06T12:12:53.635-07:00Problema 10 de la sección Nivel UNI -10<iframe allowtransparency="true" frameborder="0" height="385" scrolling="no" seamless="seamless" src="http://files.flipsnack.com/iframe/embed.html?hash=ftiydf8m&wmode=window&bgcolor=EEEEEE&t=1373137362" width="440"></iframe>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-4231777592611297753.post-4452089974028128982013-06-22T08:05:00.000-07:002013-06-22T08:06:45.740-07:00Problema GPT12 con sus soluciones y aportes.<iframe src="http://files.flipsnack.com/iframe/embed.html?hash=fhi2uyql&wmode=window&bgcolor=EEEEEE&t=1371913312" width="440" height="385" seamless="seamless" scrolling="no" frameborder="0" allowtransparency="true"></iframe>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-4231777592611297753.post-58433637561659555022013-06-22T07:11:00.000-07:002013-06-22T08:10:08.640-07:00Libro electrónico de PERÚ GEOMÉTRICO<div style="text-align: center;">
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<a href="http://www.calameo.com/books/002526971fac6032ebfea" target="_blank">Primer folleto de poblemas propuestos por Perú Geométrico</a></div>
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Unknownnoreply@blogger.com2tag:blogger.com,1999:blog-4231777592611297753.post-24463851200970335012013-05-30T15:06:00.002-07:002013-06-18T07:11:59.085-07:00Problema PG8<div dir="ltr" style="text-align: left;" trbidi="on">
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Solución al Problema PG8<br />
<a href="http://sphotos-b.ak.fbcdn.net/hphotos-ak-ash4/1000436_10200673560935355_1194657710_n.jpg" target="_blank">Por Rubén Dario Auqui</a>, <a href="http://www.facebook.com/photo.php?fbid=10200673560935355&set=a.2172993158296.124382.1050795825&type=1&theater" target="_blank">facebook de Ruben Dario</a></div>
Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-4231777592611297753.post-33021957610394799562013-05-29T06:11:00.000-07:002013-06-11T08:12:23.411-07:00Problema PG7<div class="separator" style="clear: both; text-align: center;">
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<a href="http://perugeometrico.blogspot.com/2013/05/problema-pg-7.html" rel="nofollow" target="_blank">Solución a PG7</a>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-4231777592611297753.post-24116761769060938272013-05-28T08:02:00.000-07:002013-06-11T08:12:11.790-07:00Problema PG6<div class="separator" style="clear: both; text-align: center;">
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<a href="http://perugeometrico.blogspot.com/2013/05/problema-pg-5.html" target="_blank">Solución al Problema PG6</a>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-4231777592611297753.post-15253078026244944212013-05-28T06:13:00.000-07:002013-06-11T20:31:38.953-07:00Problema PG5<div class="separator" style="clear: both; text-align: center;">
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<a href="http://perugeometrico.blogspot.com/2013/05/problema-pg-6.html" target="_blank">Solución al Problema PG5</a>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-4231777592611297753.post-85610873008718424582013-05-28T06:10:00.000-07:002013-06-11T08:11:48.935-07:00Problema PG4<div class="separator" style="clear: both; text-align: center;">
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<a href="http://perugeometrico.blogspot.com/2013/05/problema-pg4.html" target="_blank">Solución al Problema PG4</a>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-4231777592611297753.post-1621343188161453622013-05-26T13:59:00.001-07:002013-06-11T06:12:29.486-07:00Solución al Problema PG7<div class="separator" style="clear: both; text-align: center;">
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Solución <b>1</b> de Juan Felipe Crisostomo Ramos</div>
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Solución <b>2</b> de Francisco García Capitán</div>
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Solución <b>3</b> de Israel Díaz Acha</div>
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Solución <b>4</b> de Miguel Ochoa Sánchez</div>
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Solución <b>5</b> de Rubén Dario Auqui</div>
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Solución <b>6</b> de Rubén Dario Auqui</div>
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Solución <b>7</b> de Erico Freddy Palacios Loayza</div>
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<span lang="ES" style="font-family: "Arial","sans-serif"; mso-ansi-language: ES;">Solución <b>8</b> de Julio Orihuela Bastidas. </span></div>
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<span lang="ES" style="font-family: "Arial","sans-serif"; mso-ansi-language: ES;">Tomado de su facebook, dice<span style="color: #333333;">: </span></span></div>
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<span lang="ES" style="font-family: "Arial","sans-serif"; mso-ansi-language: ES;"><span style="color: #333333;">Habiendo una buena cantidad de soluciones, como que parece difícil hacer una más. Yo había procedido como Miguel Ochoa, pero la razón entre AD y DM la conseguía con semejanza (es igual a (AB)(AC)/(BM)(MC),de ahí todo lo demás es igual.<o:p></o:p></span></span></div>
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<span lang="ES" style="color: #333333; font-family: "Arial","sans-serif"; mso-ansi-language: ES;">Aquí va otra</span>:<span lang="ES" style="font-family: "Arial","sans-serif"; mso-ansi-language: ES;"><o:p></o:p></span></div>
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<span lang="ES" style="font-family: "Arial","sans-serif"; mso-ansi-language: ES;">AD/DM es igual a la razón entre la altura desde A y la flecha desde M, es decir: 2(ha)/ (Ra-r)<o:p></o:p></span></div>
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<span lang="ES" style="font-family: "Arial","sans-serif"; mso-ansi-language: ES;">análogamente los demás y se multiplica:<o:p></o:p></span></div>
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<span lang="ES" style="font-family: "Arial","sans-serif"; mso-ansi-language: ES;">(AD/DM)(BE/EN)(CF/FL)=8(ha)(hb)(hc)/(Ra-r)(Rb-r)(Rc-r)<o:p></o:p></span></div>
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<span lang="ES" style="font-family: "Arial","sans-serif"; mso-ansi-language: ES;">luego reemplazando por áreas, y considerando que: Rb-r=(S)b/p(p-b) y así los demás, se obtiene 4pp/RR<o:p></o:p></span></div>
<br />Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-4231777592611297753.post-29093504368807207522013-05-22T14:22:00.000-07:002013-06-11T08:01:19.968-07:00Solución al Problema PG 6<div class="separator" style="clear: both; text-align: center;">
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Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-4231777592611297753.post-49515035827352320682013-05-20T14:59:00.000-07:002013-06-11T08:11:38.829-07:00Solución al Problema PG5<div class="separator" style="clear: both; text-align: center;">
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<br />Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-4231777592611297753.post-10121342010115090582013-05-18T09:50:00.002-07:002013-06-11T08:17:43.444-07:00Solución al Problema PG4<div class="separator" style="clear: both; text-align: center;">
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<span style="font-family: Arial, Helvetica, sans-serif;"><b>Solución 1 por Israel Díaz</b></span></div>
<div class="UFICommentContent" id=".reactRoot[73].[0][1][1]{comment10200378895428050_2863453}.[0].[1].[0].[1].[0].[0]" style="background-color: #fafbfb; line-height: 14px;">
<span style="font-family: Arial, Helvetica, sans-serif;"><a class="UFICommentActorName" data-ft="{"tn":";"}" data-hovercard="/ajax/hovercard/hovercard.php?id=894420384&extragetparams=%7B%22hc_location%22%3A%22ufi%22%7D" href="https://www.facebook.com/israel.diaz.7923" id=".reactRoot[73].[0][1][1]{comment10200378895428050_2863453}.[0].[1].[0].[1].[0].[0].[0][0]" style="cursor: pointer; font-weight: bold; text-decoration: none;">Israel Diaz</a><span id=".reactRoot[73].[0][1][1]{comment10200378895428050_2863453}.[0].[1].[0].[1].[0].[0].[0][1]"> </span><span data-ft="{"tn":"K"}" id=".reactRoot[73].[0][1][1]{comment10200378895428050_2863453}.[0].[1].[0].[1].[0].[0].[0][2]"><span class="UFICommentBody" id=".reactRoot[73].[0][1][1]{comment10200378895428050_2863453}.[0].[1].[0].[1].[0].[0].[0][2].[0]"><span id=".reactRoot[73].[0][1][1]{comment10200378895428050_2863453}.[0].[1].[0].[1].[0].[0].[0][2].[0].[0:0]"><span id=".reactRoot[73].[0][1][1]{comment10200378895428050_2863453}.[0].[1].[0].[1].[0].[0].[0][2].[0].[0:0].[0]">Supongamos que el triangulo mayor inscrito sea ABC en donde theta es la medida del angulo B y ( E en AB ) . </span><br id=".reactRoot[73].[0][1][1]{comment10200378895428050_2863453}.[0].[1].[0].[1].[0].[0].[0][2].[0].[0:0].[1]" /><span id=".reactRoot[73].[0][1][1]{comment10200378895428050_2863453}.[0].[1].[0].[1].[0].[0].[0][2].[0].[0:0].[2]">Sea M el punto medio del segmento EF . </span><br id=".reactRoot[73].[0][1][1]{comment10200378895428050_2863453}.[0].[1].[0].[1].[0].[0].[0][2].[0].[0:0].[3]" /><span id=".reactRoot[73].[0][1][1]{comment10200378895428050_2863453}.[0].[1].[0].[1].[0].[0].[0][2].[0].[0:0].[4]">Y vamos a nombrar las variables angulares: </span><br id=".reactRoot[73].[0][1][1]{comment10200378895428050_2863453}.[0].[1].[0].[1].[0].[0].[0][2].[0].[0:0].[5]" /><span id=".reactRoot[73].[0][1][1]{comment10200378895428050_2863453}.[0].[1].[0].[1].[0].[0].[0][2].[0].[0:0].[6]">Sean alfa=m<EPB ; x=m<EBP ; phi =m<EMP . </span></span><span id=".reactRoot[73].[0][1][1]{comment10200378895428050_2863453}.[0].[1].[0].[1].[0].[0].[0][2].[0].[0:3]"><span id=".reactRoot[73].[0][1][1]{comment10200378895428050_2863453}.[0].[1].[0].[1].[0].[0].[0][2].[0].[0:3].[0]"><br id=".reactRoot[73].[0][1][1]{comment10200378895428050_2863453}.[0].[1].[0].[1].[0].[0].[0][2].[0].[0:3].[0].[0]" /><span id=".reactRoot[73].[0][1][1]{comment10200378895428050_2863453}.[0].[1].[0].[1].[0].[0].[0][2].[0].[0:3].[0].[1]">Vamos a utilizar un teorema (*) conocido de la PRE: </span><br id=".reactRoot[73].[0][1][1]{comment10200378895428050_2863453}.[0].[1].[0].[1].[0].[0].[0][2].[0].[0:3].[0].[2]" /><span id=".reactRoot[73].[0][1][1]{comment10200378895428050_2863453}.[0].[1].[0].[1].[0].[0].[0][2].[0].[0:3].[0].[3]">m<EPB=m<MPF . el cual es la llave de la solucion .</span><br id=".reactRoot[73].[0][1][1]{comment10200378895428050_2863453}.[0].[1].[0].[1].[0].[0].[0][2].[0].[0:3].[0].[4]" /><br id=".reactRoot[73].[0][1][1]{comment10200378895428050_2863453}.[0].[1].[0].[1].[0].[0].[0][2].[0].[0:3].[0].[5]" /><span id=".reactRoot[73].[0][1][1]{comment10200378895428050_2863453}.[0].[1].[0].[1].[0].[0].[0][2].[0].[0:3].[0].[6]">Empecemos m>ACP=x luego se cumple que :</span><br id=".reactRoot[73].[0][1][1]{comment10200378895428050_2863453}.[0].[1].[0].[1].[0].[0].[0][2].[0].[0:3].[0].[7]" /><span id=".reactRoot[73].[0][1][1]{comment10200378895428050_2863453}.[0].[1].[0].[1].[0].[0].[0][2].[0].[0:3].[0].[8]">PQ = PC sen(x) ......(1)</span><br id=".reactRoot[73].[0][1][1]{comment10200378895428050_2863453}.[0].[1].[0].[1].[0].[0].[0][2].[0].[0:3].[0].[9]" /><span id=".reactRoot[73].[0][1][1]{comment10200378895428050_2863453}.[0].[1].[0].[1].[0].[0].[0][2].[0].[0:3].[0].[10]">Ahora ley de senos en el triangulo BEP :</span><br id=".reactRoot[73].[0][1][1]{comment10200378895428050_2863453}.[0].[1].[0].[1].[0].[0].[0][2].[0].[0:3].[0].[11]" /><span id=".reactRoot[73].[0][1][1]{comment10200378895428050_2863453}.[0].[1].[0].[1].[0].[0].[0][2].[0].[0:3].[0].[12]">EB . sen(x) = PE . sen(alfa) .....(2) </span><br id=".reactRoot[73].[0][1][1]{comment10200378895428050_2863453}.[0].[1].[0].[1].[0].[0].[0][2].[0].[0:3].[0].[13]" /><span id=".reactRoot[73].[0][1][1]{comment10200378895428050_2863453}.[0].[1].[0].[1].[0].[0].[0][2].[0].[0:3].[0].[14]">Los triangulos PMC y PEF son semejantes : ( la prueba es usando (*) y ciclicos ) </span><br id=".reactRoot[73].[0][1][1]{comment10200378895428050_2863453}.[0].[1].[0].[1].[0].[0].[0][2].[0].[0:3].[0].[15]" /><span id=".reactRoot[73].[0][1][1]{comment10200378895428050_2863453}.[0].[1].[0].[1].[0].[0].[0][2].[0].[0:3].[0].[16]">PE . PC = PM . PF .....(3) </span><br id=".reactRoot[73].[0][1][1]{comment10200378895428050_2863453}.[0].[1].[0].[1].[0].[0].[0][2].[0].[0:3].[0].[17]" /><span id=".reactRoot[73].[0][1][1]{comment10200378895428050_2863453}.[0].[1].[0].[1].[0].[0].[0][2].[0].[0:3].[0].[18]">Por cateto opuesto a hipotenusa : </span><br id=".reactRoot[73].[0][1][1]{comment10200378895428050_2863453}.[0].[1].[0].[1].[0].[0].[0][2].[0].[0:3].[0].[19]" /><span id=".reactRoot[73].[0][1][1]{comment10200378895428050_2863453}.[0].[1].[0].[1].[0].[0].[0][2].[0].[0:3].[0].[20]">PM . sen (phi ) = PS ...(4)</span><br id=".reactRoot[73].[0][1][1]{comment10200378895428050_2863453}.[0].[1].[0].[1].[0].[0].[0][2].[0].[0:3].[0].[21]" /><span id=".reactRoot[73].[0][1][1]{comment10200378895428050_2863453}.[0].[1].[0].[1].[0].[0].[0][2].[0].[0:3].[0].[22]">MF = BF . sen(theta/2) ...(5)</span><br id=".reactRoot[73].[0][1][1]{comment10200378895428050_2863453}.[0].[1].[0].[1].[0].[0].[0][2].[0].[0:3].[0].[23]" /><span id=".reactRoot[73].[0][1][1]{comment10200378895428050_2863453}.[0].[1].[0].[1].[0].[0].[0][2].[0].[0:3].[0].[24]">Aplicando ley de senos en el triangulo PMF :</span><br id=".reactRoot[73].[0][1][1]{comment10200378895428050_2863453}.[0].[1].[0].[1].[0].[0].[0][2].[0].[0:3].[0].[25]" /><span id=".reactRoot[73].[0][1][1]{comment10200378895428050_2863453}.[0].[1].[0].[1].[0].[0].[0][2].[0].[0:3].[0].[26]">PF . sen (alfa) = MF . sen (phi) ....(6) </span><br id=".reactRoot[73].[0][1][1]{comment10200378895428050_2863453}.[0].[1].[0].[1].[0].[0].[0][2].[0].[0:3].[0].[27]" /><span id=".reactRoot[73].[0][1][1]{comment10200378895428050_2863453}.[0].[1].[0].[1].[0].[0].[0][2].[0].[0:3].[0].[28]">Y tambien por isosceles : BF=EB ....(7)</span><br id=".reactRoot[73].[0][1][1]{comment10200378895428050_2863453}.[0].[1].[0].[1].[0].[0].[0][2].[0].[0:3].[0].[29]" /><br id=".reactRoot[73].[0][1][1]{comment10200378895428050_2863453}.[0].[1].[0].[1].[0].[0].[0][2].[0].[0:3].[0].[30]" /><span id=".reactRoot[73].[0][1][1]{comment10200378895428050_2863453}.[0].[1].[0].[1].[0].[0].[0][2].[0].[0:3].[0].[31]">Ahora multiplicando (1) x (2) x (3) x ....x (7) y cancelando se obtiene que : PQ=PS sen(theta/2) .</span></span></span></span></span></span></div>
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<span style="font-family: Arial, Helvetica, sans-serif;"><b>Solución 2 por</b></span></div>
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Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-4231777592611297753.post-61914384415309614682013-05-18T09:40:00.000-07:002013-05-23T19:36:24.795-07:00Problema PG3<div class="separator" style="clear: both; text-align: center;">
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhN9ZMIUyhx5gnmUA6iIqvhttQyDD0-TNYLZBC4qOBhEMSW0fJVYyKf7KHRDyKDRBaBXUlSDDxkEMGC7XR2E0TzWr2Z3-SJIJjMsTsJlcYV17csR6X6h702TorpnTMh3vizBt6qXvblNIY/s1600/prob20.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="270" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhN9ZMIUyhx5gnmUA6iIqvhttQyDD0-TNYLZBC4qOBhEMSW0fJVYyKf7KHRDyKDRBaBXUlSDDxkEMGC7XR2E0TzWr2Z3-SJIJjMsTsJlcYV17csR6X6h702TorpnTMh3vizBt6qXvblNIY/s400/prob20.jpg" width="400" /></a></div>
<span style="font-size: large;">Solución de <span style="color: blue;">Juan Felipe Crisostomo Ramos</span></span><br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjbzrk3vSZyY14mKxqMTO5DnG7j_TI_u-HMB3bhnAz0HoTFQDS2vbFrwLeQcFghUsNnlOxnzlbO9Lp6LXLKF8jeyZDKUC3JduNnijCw_Ym08rlu-otHp0dkex2XLzicyg4-wNcHh8S4iBU/s1600/sol20..3.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="203" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjbzrk3vSZyY14mKxqMTO5DnG7j_TI_u-HMB3bhnAz0HoTFQDS2vbFrwLeQcFghUsNnlOxnzlbO9Lp6LXLKF8jeyZDKUC3JduNnijCw_Ym08rlu-otHp0dkex2XLzicyg4-wNcHh8S4iBU/s400/sol20..3.jpg" width="400" /></a></div>
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<span style="font-size: large;">Solución 1 de <span style="color: blue;">John Ascona Briceño</span></span></div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjcpxzZyGnNkqxwGcanbj3EItKI18nJ3XLv8_BpX3-wgxUgiETMg2LAmvWMIiY6RC_M9mVyOtD47q_YY97-naHT4YhryVa_BBR60WJOhGFAoyQQhYiatSjVH8IpqiZOhhBtQjU2LZELt8A/s1600/sol20..2.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="310" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjcpxzZyGnNkqxwGcanbj3EItKI18nJ3XLv8_BpX3-wgxUgiETMg2LAmvWMIiY6RC_M9mVyOtD47q_YY97-naHT4YhryVa_BBR60WJOhGFAoyQQhYiatSjVH8IpqiZOhhBtQjU2LZELt8A/s400/sol20..2.jpg" width="400" /></a></div>
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<span style="font-size: large;">Solución 2 de <span style="color: blue;">John Ascona Briceño</span></span></div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjNKsaTRbnElj0vY4Hvv0zsuYh3k-rfji4UvhQ1sbvJz4VB2kz3QZ3EA7KfLWbViUlERXYCV3H84U1dRbzhFDq_0DI9w8h0xy-LGdGUu32k9WZKLqyT8pIALWWNYsJuHY1fsqnXR3QWszU/s1600/sol20.1.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="285" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjNKsaTRbnElj0vY4Hvv0zsuYh3k-rfji4UvhQ1sbvJz4VB2kz3QZ3EA7KfLWbViUlERXYCV3H84U1dRbzhFDq_0DI9w8h0xy-LGdGUu32k9WZKLqyT8pIALWWNYsJuHY1fsqnXR3QWszU/s400/sol20.1.jpg" width="400" /></a></div>
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Solución de <span style="color: blue;">Carlos Olivera Diaz</span> :</div>
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Se demuestra que AL = AM = AD = h y al demostrar , por trigonometria, Juanfelipe que el cuadrado de h = 2AJ.AK, POR DIFERENCIA DE AREAS SE LLEGA : S1 = S2</div>
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Solución de <span style="color: blue;">Armando Ricardo Michuy Romero</span>:</div>
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<span id=".reactRoot[175].[1][4][1]{comment10200404312044301_3234261}.0.[1].0.[1].0.[0].[0][1]"> </span><span data-ft="{"tn":"K"}" id=".reactRoot[175].[1][4][1]{comment10200404312044301_3234261}.0.[1].0.[1].0.[0].[0][2]"><span id=".reactRoot[175].[1][4][1]{comment10200404312044301_3234261}.0.[1].0.[1].0.[0].[0][2].0.[0]">Si llamamos O al centro de la circunferencia menor y O" centro de la mayor, deducimos que los triángulos BNO y O"QC son semejantes, también los triángulos BMO" y APO son semejantes. De la primera y segunda semejanza se obtiene: n.m=AP.QC. </span></span></div>
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Entonces: BH al cuadrado es igual a 2.n.m.</div>
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Solución de <span style="color: blue;">Josué García Piscoya</span>:</div>
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<span data-ft="{"tn":"K"}" id=".reactRoot[175].[1][4][1]{comment10200404312044301_3234847}.0.[1].0.[1].0.[0].[0][2]">Sea [] forma de mencionar área encerrada de ; [ABH]=AN.BNcot(90º/2) esta es una propiedad que alguna vez la vi en la academia, y ya que AH.HC=BH² entonces [ABH][BHC]=(BH^4)/4 y AN/BN=BM/MC (esto es ver la semejanza) AN.MC=BN.BM=mn. Entonces BH²/2=mn ya está.</span></div>
Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-4231777592611297753.post-87000310518868766902013-05-17T15:36:00.001-07:002013-05-23T19:35:52.031-07:00Problema PG2<div class="separator" style="clear: both; text-align: center;">
<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjCxJD44tuxX0sIUdv6OHmw3TNRz0lxsoPoGPJ9Nggz-30ucYvUTgYhwHj8PZSmKKVPaRyCHV-W4vHhWQhjpPx8m8LPN8bjRFg5Z5nKaPSC3JtvZER0k_tkdSMdr-7dkTrf5XhBi4koJKs/s1600/prob21.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="269" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjCxJD44tuxX0sIUdv6OHmw3TNRz0lxsoPoGPJ9Nggz-30ucYvUTgYhwHj8PZSmKKVPaRyCHV-W4vHhWQhjpPx8m8LPN8bjRFg5Z5nKaPSC3JtvZER0k_tkdSMdr-7dkTrf5XhBi4koJKs/s400/prob21.jpg" width="400" /></a></div>
Soluciones de:<br />
Miguel Ochoa Sánchez<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgKu4ur9coTmPyg80PTuZ2Ll-T5VqQs5VdtWU4xyBJ1RF0tMQfmmy2ImjjcE4g7eaEUFi1Q3hFwLMNifGgLpkB5e3YgUrJzLcxUi1eCX5gmzNPFhAdGP6iQT56PYwVIgSaFbYhvfkN7J1s/s1600/sol21.1.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="356" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgKu4ur9coTmPyg80PTuZ2Ll-T5VqQs5VdtWU4xyBJ1RF0tMQfmmy2ImjjcE4g7eaEUFi1Q3hFwLMNifGgLpkB5e3YgUrJzLcxUi1eCX5gmzNPFhAdGP6iQT56PYwVIgSaFbYhvfkN7J1s/s400/sol21.1.jpg" width="400" /></a></div>
Solución1 de Rubén Darío Auqui<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjU-MQxE8eogbMMfr6xa2wQ0lp0AMp887QDoTaX1tFbrGwBEl5ajm4Z2iksnfY57rkouNrsUCCSEgIigwnpIGMZIx_O1CNXRv8crrHS1gyp7VLdKXlVU6p2w1VPBP5MKxmh4LBWvtsabCM/s1600/sol21.2.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="300" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjU-MQxE8eogbMMfr6xa2wQ0lp0AMp887QDoTaX1tFbrGwBEl5ajm4Z2iksnfY57rkouNrsUCCSEgIigwnpIGMZIx_O1CNXRv8crrHS1gyp7VLdKXlVU6p2w1VPBP5MKxmh4LBWvtsabCM/s400/sol21.2.jpg" width="400" /></a></div>
Solución de Paul Antonio Huamaní López<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjYREyhLkcyLDv941hKOuVISnUuRuo8zqYb6cHa0oaxQv4NZiPszvGuAzTfQXD-mh7ZXnQX922Qh9U5-J3qZhD0ZeN9iV3SksAWMDO3wAvbsTmC2LZGYejup56AIf5P3TVYO66VAkWOpQo/s1600/sol21.4.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="222" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjYREyhLkcyLDv941hKOuVISnUuRuo8zqYb6cHa0oaxQv4NZiPszvGuAzTfQXD-mh7ZXnQX922Qh9U5-J3qZhD0ZeN9iV3SksAWMDO3wAvbsTmC2LZGYejup56AIf5P3TVYO66VAkWOpQo/s400/sol21.4.jpg" width="400" /></a></div>
<br />Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-4231777592611297753.post-84671300735327732932013-05-17T15:36:00.000-07:002013-05-23T19:35:17.647-07:00Problema PG1<div class="separator" style="clear: both; text-align: center;">
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjjAz4Ufp3Ocdqq-ZUqviz_PJ_Zv9LWDOKFOvWiQ2DvXxERIHey5mmGlveyCAsj9Ir5__A56kyFMPxZluSzBWrwTDivOfwhhZhtd12A5SBf0GxIs5ycWP8Xqs2_79vLb53xfhG9wQf1jU4/s1600/prob22.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="244" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjjAz4Ufp3Ocdqq-ZUqviz_PJ_Zv9LWDOKFOvWiQ2DvXxERIHey5mmGlveyCAsj9Ir5__A56kyFMPxZluSzBWrwTDivOfwhhZhtd12A5SBf0GxIs5ycWP8Xqs2_79vLb53xfhG9wQf1jU4/s400/prob22.jpg" width="400" /></a></div>
Soluciones:<br />
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Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-4231777592611297753.post-30191275775739706152013-05-17T00:30:00.000-07:002013-05-30T12:43:51.264-07:00Problema PG de inaguración PG - 0<div class="separator" style="clear: both; text-align: center;">
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi6Qu2txMKeXabnpqglrm7szgnVqFAReqrPB0HaiVh7Q6kP_yrjvpNmzpmG5W94JkWtrTs6wLcpyBVYS2KTmBm7xUgnJaBnoFDqqFe8Z_d07LM9O3kP3gmefzorvPNozSKPsgldQvxrnmE/s1600/erico+freddy+palacios+loayza+8.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="266" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi6Qu2txMKeXabnpqglrm7szgnVqFAReqrPB0HaiVh7Q6kP_yrjvpNmzpmG5W94JkWtrTs6wLcpyBVYS2KTmBm7xUgnJaBnoFDqqFe8Z_d07LM9O3kP3gmefzorvPNozSKPsgldQvxrnmE/s400/erico+freddy+palacios+loayza+8.png" width="400" /></a></div>
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Solución <span style="color: red;">(1)</span> de Israel Díaz Acha: </div>
<span id=".reactRoot[191].[1][4][1]{comment10200162976051052_3130103}.0.[1].0.[1].0.[0].[0][1]"> </span><span data-ft="{"tn":"K"}" id=".reactRoot[191].[1][4][1]{comment10200162976051052_3130103}.0.[1].0.[1].0.[0].[0][2]">Sale que :</span><br />
<span data-ft="{"tn":"K"}">[A1A2A3]=(rArBrC)^2.sen(1/2(A-B))sen(1/2(B-C))sen(1/2(C-A))/8sen(1/2A)sen(1/2B)sen(1/2C) . Y de allí con sus simplificaciones trigonométricas respectivas se debe conseguir dicha igualdad .</span><br />
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<span data-ft="{"tn":"K"}">Solución <span style="color: red;">(2)</span> de Juan Felipe Crisostomo Ramos</span></div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhOYASyGDW1gXpVdP_ZMgFgtCHIZ3PYgUZF8f3uqAQL9neC_BUursXHrPlVad5d2h63M2zetnWKvn7A5xOOFATeWbarWWiSBFTduvY-gSo3SpUW74gArlxgkDOCXME9Vvn09wo5TokPl24/s1600/juan%252Bfelipe%252Bcrisostomo%252Bramos.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="176" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhOYASyGDW1gXpVdP_ZMgFgtCHIZ3PYgUZF8f3uqAQL9neC_BUursXHrPlVad5d2h63M2zetnWKvn7A5xOOFATeWbarWWiSBFTduvY-gSo3SpUW74gArlxgkDOCXME9Vvn09wo5TokPl24/s400/juan%252Bfelipe%252Bcrisostomo%252Bramos.png" width="400" /></a></div>
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Solución <span style="color: red;">(3)</span> de Julio Orihuela Bastidas</div>
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjzo_-z5MTk-VdwyBjhTUDjb50mGiH0YASJ7ZVO8K8gUis7OnQwsQ8o9yLwZOrlzxEDfhilewdCMXBR6irRBOLk2WSj5DRyoT8erKthhjdpUGD9CbrOkjJBMWjd1hpIpLRItHnNXC5h5IY/s1600/julio+orihuela+bastidas.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="400" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjzo_-z5MTk-VdwyBjhTUDjb50mGiH0YASJ7ZVO8K8gUis7OnQwsQ8o9yLwZOrlzxEDfhilewdCMXBR6irRBOLk2WSj5DRyoT8erKthhjdpUGD9CbrOkjJBMWjd1hpIpLRItHnNXC5h5IY/s400/julio+orihuela+bastidas.png" width="265" /></a></div>
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Solución <span style="color: red;">(4)</span> de Carlos Olivera Días:</div>
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<span lang="ES" style="color: #333333; mso-ansi-language: ES;">Carlos
Olivera Díaz, saluda y felicita a todos los participantes en el presente
problema que tiene una bonita presentación, pareciera un problema difícil, sin
embargo, el área de cada triangulo se puede calcular multiplicando cada radio
exinscrito correspondiente, por la respectiva proyección de la ceviana de NAGEL</span>
ENTRE 2 Y LUEGO MULTIPLICANDO ESTOS TRES RESULTADOS se llega a la relación
propuesta.<o:p></o:p></div>
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Solución <span style="color: red;">(5)</span> de Erico Freddy Palacios Loayza</div>
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Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-4231777592611297753.post-89568846618094886212013-05-16T23:28:00.000-07:002013-11-09T16:06:28.755-08:00PRESENTACIÓN<h5 class="uiStreamMessage userContentWrapper" data-ft="{"type":1,"tn":"K"}" style="text-align: justify;">
<span class="messageBody"><span class="userContent"><span style="color: #333333;">PERÚ GEOMÉTRICO NACE CON EL OBJETIVO DE FOMENTAR Y ALENTAR EL DESARROLLO DE LA GEOMETRÍA EN SUS DIFERENTES NIVELES.<br />NUESTRO COMITÉ EDITORIAL ESTÁ CONFORMADO POR UN CONJUNTO DE AFICIONADOS A LA GEOMETRÍA QUE CON SU AMPLIA TRAYECTORIA Y EXPERIENCIA, DESEA APORTAR Y LLEGAR A TODOS USTEDES Y MOSTRARLES EL ARTE Y BELLEZA QUE ESTA CIENCIA ABARCA.</span></span></span><span class="messageBody"><span class="userContent"><span style="color: #333333;"><br />ASÍ MISMO QUEREMOS EXPRESAR NUESTRO MÁS PROFUNDO AGRADECIMIENTO Y RECONOCIMIENTO TANTO EN EL PERÚ Y EN EL EXTRANJERO A TODAS AQUELLAS PERSONAS QUE NOS HAN MOTIVADO Y ALENTADO INCESANTEMENTE EN LA ELABORACIÓN DE ESTE BLOG.</span></span></span></h5>
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COMITÉ EDITORIAL<o:p></o:p></div>
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Integrantes del Comité Editorial
de PERÚ GEOMÉTRICO:<br />
<ul>
<li><a href="https://www.facebook.com/ruben.dario.7315720">Auqui Cáceres Rubén</a></li>
<li><a href="http://www.facebook.com/#%21/juanfecris.ramos?fref=ts">Crisostomo Ramos Juan</a></li>
<li><a href="http://www.facebook.com/#%21/edson.curahuaortega?fref=ts">Curahua Ortega Edson</a></li>
<li><a href="https://www.facebook.com/PasaPaca.Snack?hc_location=stream">García Piscoya Josué</a></li>
<li><a href="http://www.facebook.com/#%21/RMCESC?fref=ts">Mariños Castillo Roberto</a></li>
<li><a href="http://www.facebook.com/#%21/miguel.ochoasanchez.9">Ochoa Sánchez Miguel</a></li>
<li><a href="http://www.facebook.com/julio.orihuela.39?ref=ts&fref=ts">Orihuela Bastidas Julio</a></li>
<li><a href="https://www.facebook.com/erikferpo">Palacios Loayza Erico</a></li>
<li><a href="http://www.facebook.com/julio.orihuela.39?ref=ts&fref=ts#%21/manuel.turjillovillegas?fref=ts">Trujillo Villegas Manuel</a></li>
</ul>
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