Libro 2014
Short Description
Campeonato de Matemática - 2014...
Description
PROBLEMAS Y SOLUCIONES 7◦ Camp Campeo eona nato to de Matem´ Mat em´ atica atica Universidad de La Frontera
Departamento de Matem´ atica ati ca y Estad Est ad´´ıstica ıst ica Universidad de La Frontera
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El objetivo principal del matem´atico atico consiste en formular y resolver problemas “La resoluci´on on de problemas es el coraz´ on on de la Matem´atica” atica”
Soluciones de los Problemas a cargo del Comit´ Com it´e Acad´ Ac ad´emico emi co del Campe Cam peona onato to Con Confor formado mado po por: r: Hern´an an Burgos Vega Jos´e Labr Labrin in Parra Par ra Eduardo Milman Aguilar
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Introducci´ on on Este peque˜no no libro contiene los problemas del 7 Campeonato de Matem´atica atica de la Universidad de La Frontera a˜no no 2014, junto a sus soluciones, material que ponemos a disposici´on on de cada uno de los colegas con la ´ıntima esperanza esp eranza y pretensi´ p retensi´on on de que les sirvan de ayuda y motivaci´on en sus clases de matem´atica. atica. ◦
Estamo Estamoss conv convencido encidoss de que estudi estudian antes tes entus entusias iasmad mados os y motiv motivados ados tendr´an an mayor mayor ´exito, exito, como tambi´ tambi´en, en, pensamos que problemas no triviales viales y que desaf´ desaf´ıan el intelec intelecto to del estudian estudiante te lo ayudan ayudan a organizar, organizar, contar, razonar, encontrar secuencias y algoritmos que le permiten entender los problemas y resolverlos, y al final del proceso sentir el placer de haber hab er resuelto un problema probl ema novedoso, que inicialmente inicia lmente era un desaf´ desaf´ıo. Este libro est´a dirigi dirigido do especi especialm almen ente te a los colega colegass profe profesor sores es de matem´atica atica de ense˜nanza nanza b´asica a sica y media de la regi´on on de La Araucan´ıa, ıa, con el prop´osito osito de despertar el inter´ inter´es es por la resoluci´on on y creaci´on o n de problemas l´ udicos y bonitos que motivan a nuestros estudiantes. udicos Quiero agradecer agrad ecer el importante impo rtante aporte apo rte del de l comit´ co mit´e acad´ a cad´emico emico del campeocamp eonato, en la resoluci´on on de los problemas seleccionados, adem´as as de agradecer por la lectura y correcciones del material a la profesora Elena Olivos del Departamento de Matem´atica atica de nuestra Universidad. Debemos reconocer reconocer tambi´ tambi´ en en el esforzado esforzado y excelen excelente te trabajo desarrollad desarrolladoo por el profesor Jos´e Labrin en la escritura del texto en LaTeX. LaTeX. Finalmente, es necesario destacar y reconocer el importante y decidido apoyo que nos entrega la direcci´on on de nuestra Universidad, sin el cual no podr po dr´´ıamos desarrollar desarro llar este Campeon Ca mpeonato. ato. Reiteram Reiteramos os nuestro nuestro ´ıntimo ıntimo deseo deseo de que este material sea un peque˜no no aporte a los colegas en su pr´actica actica diaria. Dr. Hern´an an Burgos V. Director Acad´emico emico Campeonato Campeo nato de Matem´atica atica Universidad de La Frontera
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1 Problemas
e dibujo es la parte central Problema 1. ¿Qu´ de la estrella que se muestra en la imagen ?
(A)
(B)
(C )
(D)
(E)
Problema 2. Juan quiere insertar el d´ıgito 3 en alg´ un un lugar del n´ umero umero
2014. ¿D´onde onde se debe inserta insertarr el d´ıgito ıgito 3 si el quiere quiere que su n´ umero de cinco cin co d´ıgitos ıgi tos sea lo m´ as as peque˜ no no posible? Problema 3. ¿Qu´e pareja de casas fueron hechas usando exactamente las
mismas piezas de forma triangular o rectangular?
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2
Al icia tiene t iene 10 1 0 nietos. niet os. Fernanda es la mayor. Un d´ d´ıa la abuela Problema 5. Alicia se da cuenta que sus nietos tienen todos edades distintas. Si la suma de las edades de sus nietos es 180, ¿Cu´al al es la edad m´ınima que Fernanda puede tener?
Problema 6. Mar´ Mar´ıa realiza rea liza 6 restas resta s y obtiene obt iene como co mo resultados resu ltados los n´ numeros ´
desde el 0 al 5. Ella une los puntos de menor a mayor, comenzando en el punto con el resultado 0 y terminando en el punto con el resultado 5. ¿Cu´ al al figura ella obtiene?
(A)
(B)
(C)
(D)
(E)
o menos castillos castillos de arena que Mart´ Mart´ın pero Problema Problema 7. Alan construy´ m´as as que Susana. Luc´ Luc´ıa contruy´ o m´as as castillos castillos que Alan y que Mart´ Mart´ın. Daniela construy´o m´as as cast castil illo loss que q ue Mart Mar t´ın y menos men os que Luc´ıa. ıa. ¿Qui´ ¿Qu i´en en de ellos contruy´o m´ as as castillos de arena?
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ENUNCIADOS DE LOS PROBLEMAS
3
Problema 9. Ana tiene cuatro piezas las
cuales se muestran en la imagen. Con esas piezas piezas ella puede puede cubrir cubrir comple completame tament ntee solo una de las siguientes figuras. ¿Cu´al al de las siguientes ella puede cubrir con todas sus piezas? (A)
(B)
(C)
Problema 10. Bruno ha pintado flores en la ven-
tana de la tienda (mire la figura). ¿C´omo omo se ver´an an estas flores desde el otro lado de la ventana?
(D)
(E)
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4
(A)
(B)
(C)
(D)
(E)
Problema 13. Paula dispara flechas al objeti-
vo que se muestra en la figura. Cuando ella no acierta al objetivo, obtiene cero puntos. Paula dispara dos flechas y suma el puntaje de ambos disparos. ¿Cu´al al de las siguientes sumas no puede ser su puntaje? (A)60
(B)70
(C)80
(D)90
(E)100
Mar´ıa ten´ıa ıa el mismo mis mo n´umeumeProblema Problema 14. Mar´ ro de fichas grises, negras y blancas. Ella utiliz´o algunas de estas fichas para hacer una pila. En la figura, se pueden ver todas las fichas que utiliz´o. o. Ella todav´ todav´ıa tiene cinco fichas que no est´an an en la pila. ¿Cu´antas antas fichas negras negra s ten´ ten´ıa al principio? Problema Problema 15. A un conejo le gustan mucho las zanahorias y el repollo.
En un d´ıa se puede comer o 9 zanahorias, zanahorias, o 2 repollos, o 4 zanahorias zanahorias y 1
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ENUNCIADOS DE LOS PROBLEMAS
(A)
(B)
5
(C)
(D)
(E)
antos puntos hay en esta imagen? antos Problema 17. ¿Cu´
no no tiene 20 cangurocanguroProblema 18. ¿En el planeta canguro cada canguro-a˜ meses y cada canguro-mes tiene 6 canguro-semanas. ¿Cu´antas antas canguro-
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6
o cartas ca rtas en l´ l´ınea como c omo se s e muestra muest ra en la figura. fi gura. Problema 20. Evelyn orden´ En cada movimiento a Evelyn se le permite intercambiar cualquier par de cartas. ¿Cu´ al a l es el menor n´ umero de movimientos que Evelyn tiene que umero hacer para conseguir la palabra ALOCADOS?
angulos de diamantes. En la angulos Problema 21. Se realiza una secuencia de tri´ figura se muestran las tres primeras etapas. En cada etapa se a˜ nade nade una l´ınea de diamantes. diamantes . En las l´ıneas inferiores inferi ores los diamantes exteriores exter iores son de color blanco. blanco. El resto de los diamantes diamantes en el tri´ angulo angulo son negros. ¿Cu´antos antos diamantes negros tiene la figura en la etapa n´ umero umero 6?
Problema Problema 22. Un canguro compr´ o juguetes y le entreg´o 150 canguro-
monedas al asistente de la tienda. Recibi´ o 20 canguro-monedas de vuelta. Luego cambi´o de opini´on on y cambi´o uno de los juguetes por otro. Le de¿Con
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ENUNCIADOS DE LOS PROBLEMAS
7
umeros umeros 0, Problema 23. Escribe cada uno de los n´ 1, 2, 3, 4, 5, 6 en los cuadrados para hacer la adici´on on corr co rrec ecta ta.. ¿Qu´e d´ıgit ıg itoo esta estar´ r´ a en el cuadrado gris?
Problema Problema 24. ¿Cu´ al al es el mayor n´ umero umero de
cuadrados peque˜ nos que pueden ser sombreanos dos para que ning´un un cuadrado de la forma:
Hech Hechoo de cuatr cuatroo cuad cuadrad rados os som sombrea breado doss peque˜ nos aparezca en la figura? nos Problema 25. Nicol ha escrito cada uno de los
n´umeros umeros del 1 al 9 en las celdas de la tabla 3 3. Solo cuatro de estos n´ umeros umeros se pueden ver en la figura. Nicol se ha dado cuenta de que para el n´umero umero 5 la suma de los n´ umeros umeros en las celdas vecinas es igual a 13 (las celdas vecinas son aquellas celdas que comparten lados). Se dio cuenta que lo mismo se aplica para el n´ umero umero 6. ¿Qu´e n´umero umero ha escrito Nicol en la celda
×
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8
(A) a < b < c (B) a < c < b (C) b < a < c (D) b < c < a (E) c < b < a Problema 28. ¿Q ¿Qu´e numero u ´mero est´ a en el medio de
2 4 y ? 3 5
umero umero 2014 el ultimo u ´lt imo d´ıgito ıgi to es m´as as grande que la Problema Problema 29. En el n´ suma de los otros tres d´ıgitos. ¿Cu´ antos antos a˜ nos nos atr´as a s fue la ultima u ´ ltima vez que ocurri´ o esto? Laa lon longitu gitud d de los los lad lados Proble Problema ma 30 30.. L del del hex´ hex´ agono agono regul regular ar gran grande de es do doss vece vecess la long longit itud ud de los los lado ladoss del del hex´ hex´ agon agonoo regular peque˜ no. no. El hex´agono agono peque˜ no no tiene una superficie de 4 cm2. ¿Cu´al al es el area ´area del hex´agono agono grande? grande? al al es la negaci´on on de la siguiente declaraci´on on ”Todo el Problema Problema 31. ¿Cu´
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ENUNCIADOS DE LOS PROBLEMAS
9
no no (2014) Problema 35. Una abuela, su hija y su nieta pueden decir este a˜ que la suma de sus edades edades es 100. ¿En qu´ qu´e a˜ no no naci´o la nieta si cada una de las edades es una potencia de 2?
Problema Problema 36. Pablo puso algunos cua-
dros rectangulares en la pared. Por cada foto puso un clavo en la pared 2,5 m por encima del suelo y adjunt´o una larga cadena de 2 m en las dos esquinas superiores. ¿Cu´al al de las siguientes fotos es la m´ as as cercana al suelo (formato: ancho en cm altura en cm)?
×
(A)60
× 40
(B) 120
× 50
(C) (C) 120 120
× 90 (D)160 × 60 (E)160 × 100
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10
Problema 40. En la imagen hay un dado especial.
Los n´ umeros en las caras opuestas siempre suman umeros lo mismo. Los n´umeros umeros que no podemos ver en la imagen son todos n´ umeros umer os primos pri mos.. ¿Qu´e n´ umero umero es opuesto al 14?
Problema 41. Ana ha caminado 8 kil´ ometros ometros con una velocidad velocidad de 4km/h. 4km/h.
Ahora ella correr´a alg´ un tiempo con una velocidad de 8 km/h. ¿Cu´anto un anto tiempo le queda por correr para que su velocidad promedio global sea de 5 km/h?
Problema 42. Un jugador de ajedrez jug´ o 40 partidos y acumul´o 25 pun-
tos (una victoria cuenta como un punto, un empate cuenta como medio punto, y una derrota cuenta como cero puntos). ¿Cu´ al al es la diferencia
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ENUNCIADOS DE LOS PROBLEMAS
Proble Problema ma 46 46.. En la imagen que se mues-
tra, Karina quiere a˜ nadir algunos segmentos de nadir l´ınea de tal manera que cada uno de los siete puntos pun tos tenga el mismo n´ umero umero de conexiones a los otros puntos. ¿Cu´ al al es el menor n´ umero umero de segmentos de l´ınea que Karina debe dibujar? Problema 47. La imagen mues-
tra el mismo cubo desde dos puntos de vista diferentes. Est´a construido a partir de 27 cubos peque˜ nos, nos, algunos de ellos son de color negro y algunos son blancos.
11
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12
Proble Problema ma 51. En la imagen, P T es tangen-
te a un unaa circ circun unfe fere renncia C con centro O y P B bisectriz del angulo a´ngulo T P A. Calcula el ´ anguangulo T BP .
umeros umeros de 7 d´ıgitos ıgit os Problema Problema 52. Considere el conjunto de todos los n´ que se pueden obtener utilizando, para cada n´umero, umero, todos tod os los d´ıgitos ıgito s 1, 2, 3, ..., 7. Enumere los n´umeros umeros de la serie en orden creciente y divida la lista exactamente en la mitad en dos partes de igual tama˜ no. no. ¿Cu´al al es el ultimo u ´ ltimo n´ umero de la primera mitad? umero
ABC AB C
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ENUNCIADOS DE LOS PROBLEMAS
13
antos antos cuadril´ateros ateros de Problema 56. ¿Cu´ cualquier tama˜ no se muestran en la figuno ra?
al es el resultado de 2014 2014 al Problema 57. ¿Cu´
·
÷ 2014 − 2014?
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14
o el cuerpo Problema 62. Jorge construy´ que se muestra a continuaci´ on on usando siete unidades c´ ubicas. ubicas. ¿Cu´antos antos cubos tiene que agregar para hacer un cubo con aristas de longitud 3?
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ENUNCIADOS DE LOS PROBLEMAS
Proble Problema ma 67 67.. Cinco Cinco rect´ rectangulo a´nguloss iguale igualess se
colocan dentro de un cuadrado de 24 cm de lado, como se muestra en el diagrama. ¿Cu´al al es el area a´rea de uno de los rect´ angulos angulos resultantes?
Proble Problema ma 68. El El coraz coraz´ o´n y la flecha
15
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16
an Sparrow y su tripulaci´on an on pirata desenterraron Problema 71. El Capit´ varias monedas monedas de oro. Ellos dividen dividen las monedas monedas entre entre s´ı de manera que cada persona recibe el mismo n´ umero de monedas. Si hubiera cuatro piumero ratas menos en la tripulaci´ on, on, entonces ento nces cada persona per sona recibir recibi r´ıa 10 1 0 monedas mo nedas m´as. as. Sin embargo, si hubiera 50 monedas menos, cada persona recibir´ recibir´ıa 5 monedas menos. ¿Cu´antas antas monedas desenterraron?
umeros umeros positivo positivoss es 30 % menos menos que Problema 72. El promedio de dos n´ uno de ellos. ¿En qu´e porcentaje p orcentaje es mayor este promedio que el otro otr o n´ nume´
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ENUNCIADOS DE LOS PROBLEMAS
17
Mar´ıa compiten compiten en la resoluci´ resoluci´ on on de problemas. A Problema Problema 76. Liz y Mar´ cada una de ellas se les da la misma lista de 100 problemas. La primera en resolver cualquiera de estos problemas obtiene 4 puntos, mientras que la segunda en resolverlo obtiene 1 punto. Liz resolvi´ o 60 problem pro blemas, as, y Mar´ Mar´ıa tambi´ ta mbi´en en reso resolv lvi´ i´o 60 problemas. Juntas, consiguieron 312 puntos. ¿Cu´antos antos problemas fueron resueltos por ambas? ´ Problema 77. David viaja en su bicicleta desde Temuco a su parcela. El deb´ deb´ıa llegar a las 15 : 00, pero gast´ o 32 del tiempo planeado cubriendo
3 4
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18
Problema Problema 81. Una plaza de 5
× 5 est´a he-
cha de 25 azulejos de 1 1, todos los azulejos con el mismo patr´on, on, tal como el azulejo que
×
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ENUNCIADOS DE LOS PROBLEMAS
22014 Problema 86. Calcule el valor de 2013 2
19 2013
−2 −2
2012
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20
umero umero 2014 los d´ıgitos son diferentes diferentes y el ultimo u´ltimo Problema Problema 91. En el n´ d´ıgito ıgit o es mayor que q ue la suma de los otros tres d´ıgitos ıgit os ¿Cu´ antos antos a˜ nos nos antes de 2014 ocurri´ o esto por ultima u ´ ltima vez?
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ENUNCIADOS DE LOS PROBLEMAS
21
a, b, c son n´ umeros umeros no nulos, n es un n´ umero umero entero positivo. Problema 96. a, Se sabe que los n´ umeros umeros ( 2)2n+3 a2n+2 b2n−1 c3n+2 y ( 3)2n+2 a4n+1 b2n+5 c3n−4 tienen el mismo signo. ¿Cu´ al de las siguientes alternativas es al
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22
es un rect´angulo. angulo. Problema 102. PQRS es
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58
a´rea del paraleloProblema Problema 58. El area gramo ABCD es 10. Los puntos M y N son puntos medios de los lados AD y BC . ¿Cu´ al a l es el area a´rea del cuadril´atero atero MBND ? Soluci´on on Al unir los puntos M y N se forman cuatro tri´angulos angulos equivalentes (de igual area), ´area), por lo tanto el ´area area de cada tri´angulo angulo es 2.5, finalmente el ´area area del cuadril´atero atero sombreado MBND es 5. Problema 59. El producto de dos n´ umeros es 36 y la suma es 37. ¿Cu´al umeros al
es la diferencia positiva entre ellos? Soluci´on on Sean x y 37 x los n´ umeros buscados, por lo tanto x (37 x) = 36, umeros al resolver esta ecuaci´ on on cuadr´atica atica se obtienen como soluciones 1 y 36. Al considerar x = 1 = 37 x = 36, al considerar x = 36 = 37 x = 1 y en ambos casos la diferencia positiva es 35.
−
⇒ −
· − ⇒ −
area area 4. Ella Problema Problema 60. Amanda tiene varios cuadrados de papel de ´ los corta en cuadrados y en tri´ angulos angulos rect´angulos angulos de la forma en que se muestra en el primer diagrama. Luego toma algunas de las piezas y hace un p´ajaro ajaro como se muestra en el segundo diagrama. ¿Cu´al a l es el area ´area de este p´ ajaro? ajaro?
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SOLUCIONES DE LOS PROBLEMAS
59
Como el ´area area del cuadrado original es 4 y la diagonal lo divide en dos partes iguales, el tri´ angulo angulo grande tiene area a´rea 2, el ´area area del cuadrado peque˜ no n o es 1 y el ´area area de cada uno de los tri´ angulos angulos peque˜ nos nos tiene area ´area 21 . Identificando cada una de estas partes en la nueva figura notamos que su area ´area es 6. Problema 61. Un estanque estaba estaba lleno hasta su mitad, el abuelo AnacleAnacle-
to a˜ nadi´ nadi´o 2 litros al estanque. Ahora el estanque est´a lleno a tres cuartos de su capacidad. ¿Cu´al al es la capacidad del estanque? Soluci´on on Si el estanque estaba hasta la mitad y con 2 litros adicionales qued´o a tres cuartos de su capacidad, significa que los dos 2 litros que a˜nadi´ nadi´o el 1 abuelo corresponden a 4 de la capacidad. Luego, 2 =
capacidad
4
⇒ capacidad = 2 · 4 = 8 litros
Problema 62. Jorge construy´ o el cuerpo
que se muestra a continuaci´ on on usando siete unidades c´ ubicas. ubicas. ¿Cu´antos antos cubos tiene que agregar para hacer un cubo con aristas de longitud 3?
Soluci´on on Como el cubo que se debe formar es de arista 3, este contendr´a 27 cubos
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60
Soluci´on on Notemos Notemos que: 44
× 777 = 11 × 4 × 111 × 7 = 11 × 111 × 4 × 7 55 × 666 = 11 × 5 × 111 × 6 = 11 × 111 × 5 × 6 77 × 444 = 11 × 7 × 111 × 4 = 11 × 111 × 7 × 4 88 × 333 = 11 × 8 × 111 × 3 = 11 × 111 × 8 × 3 99 × 222 = 11 × 9 × 111 × 2 = 11 × 111 × 9 × 2 Observemos que todos los n´ umeros tienen los factores 11 111 , por lo umeros que analizamos los otros dos factores, factores, donde 5 6 es el mayor, por lo tanto, el resultado m´as as grande es 55 666.
×
×
×
Problema 64. El collar de la imagen contiene perlas grises y perlas blan-
cas. Marcos saca una perla tras otra del collar. Siempre saca una perla de uno de los extremos. Se detiene cuando ha quitado la quinta perla gris. ¿Cu´ al al es el mayor n´ umero de perlas blancas que Marcos puede haber saumero cado?
Soluci´on on El mayor n´ umero de perlas blancas que Marcos puede sacar es 7, lo cual umero se deduce de la siguiente manera, Marcos quita una perla gris de cada lado sacando luego 3 blancas, Marcos quita nuevamente una perla gris de cada lado y puede sacar 4 blancas m´as, as, finalmente Marcos quita la quinta perla gris de cualquiera de los lados del collar, se detiene y no puede quitar m´as as perlas. Finalmente el mayor n´ umero de perlas blancas que Marcos puede umero
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SOLUCIONES DE LOS PROBLEMAS
61
Soluci´on on Observemos que por cada lecci´ on de Alejandra Juan toma 4 lecciones, on por lo que la cantidad de lecciones que toma Alejandra y la cantidad de lecciones que toma Juan est´ an an en la raz´on on 41 . Sea x la cantidad de lecciones que ha tomado Alejandra queremos que Juan haya tomado x + 15 lecciones, por lo que debe ocurrir que: x 1 = 4 x + 15 Resolviendo la ecuaci´on on se tiene x = 5, o sea Alejandra ha tomado 5 lecciones, es decir han transcurrido 10 semanas, por lo tanto Juan ha tomado 20 lecciones. Problema 66. En el diagrama,
el area a´rea de cada c´ırculo es 1cm2. El area a´rea com´ un a dos c´ırculo un ırc uloss superpuestos es 81 cm2. ¿Cu´al a l es el area ´area de la regi´ on on cubierta por los cinco ci nco c´ırcul ır culos os?? Soluci´on on El area a´rea de cada c´ırculo es 1 cm2, entonces, el area ´area de 5 de ellos es 5cm2, 1 pero como dos de estos tienen un sector com´ un un de cm2, y dichos sectores 8 son 4 se tiene que el ´area area de la figura es: 5 .
− 4 · 18 = 29 cm
2
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62
Soluci´on on
Sean x e y los lados del rect´angulo, angulo, como muestra la figura, de este modo se tiene que 2x + 2y = 24 y que 3y + x
− x = 24
Obteniendo de la segunda ecuaci´ on on que y = 8, y reemplazando en la primera se tiene que x = 4, por lo que el area a´rea del rect´ angulo angulo es 8 4 = 32cm2
·
El coraz coraz´ o´n y la flecha Proble Problema ma 68. El est´an an en las posiciones mostradas en la figura. Al mismo tiempo, el coraz´on o n y la flecha empiezan a moverse. La flecha se mueve tres lugares hacia la derecha y el coraz´on on se mueve cuatro lugares hacia la izquierda y luego se detiene. Siguen la misma rutina rutina una y otra vez. vez. ¿Desp ¿Despu´ u´es es de cu´antas antas veces repetida la rutina se encontrar´ a el coraz´on o n y la flecha en el mismo tri´angulo angulo por primera vez?
Soluci´on on
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SOLUCIONES DE LOS PROBLEMAS
63
Problema Problema 69. La figura muestra el tri´angulo angulo ABC en donde BH es una altura y AD es bisectriz del angulo a´ngulo en A. El angulo ´angulo obtuso entre BH y AD es cuatro veces el angulo a´ngulo DAB . ¿Cu´anto anto mide el angulo a´ngulo C AB ?
Soluci´on on AE H = 90◦ α, adem´ AE H + ∠AEB AE B = 180◦, es Notemos que ∠AEH as, as, ∠AEH decir, 90◦ α + 4 α = 180◦ por lo tanto α = 30◦.
−
−
angulo tiene lados de longitudes 6 cm y 11 cm. Se angulo Problema 70. Un rect´ selecciona uno de los lados largos y se dibujan las bisectrices de los ´angulos angulos en cada extremo de ese lado. Estas bisectrices dividen el otro lado largo en tres partes. ¿Cu´ ales son las longitudes de estas secciones? ales Soluci´on on Al traz trazar ar las las bise bisect ctri rice cess como como se muestr estraa en la fig figu ura se ge
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64
m´as. as. Sin embargo, si hubiera 50 monedas menos, cada persona recibir´ recibir´ıa 5 monedas menos. ¿Cu´antas antas monedas desenterraron?
Soluci´on on Sea x el n´ umero de monedas por tripulantes, y el n´ umero umero umero de tripulantes. Entonces x y ser´a el n´ umero total de monedas. Con los datos se puede umero plantear el siguiente sistema de ecuaciones:
·
(y
− 4) · (x + 10) = x · y y · (x − 5) = x · y − 50
Dicho sistema tiene como soluci´ on x = 15 e y = 10. Finalmente se on desenterraron x y = 150 monedas.
·
umeros umeros positivo positivoss es 30 % menos menos que Problema 72. El promedio de dos n´ uno de ellos. ¿En qu´e porcentaje p orcentaje es mayor este promedio que el otro otr o n´ nume´ ro?
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SOLUCIONES DE LOS PROBLEMAS
65
Por otra parte tenemos que: x+y
3 − y 2 10 y x 3 − y + y = − 2 10 2 = y
x 1 y= 5 2 5 y= x 2
Entonces el promedio es: x+y
x + 25 x
5 + x 2 2 2 4 Como queremos saber la diferencia entre el promedio y x sumamos y restamos x a esta expresi´ on on obteniendo: =
=
x
x 5 3 + x = x + x 2 2 4 2 4 Finalmen Finalmente te el promedio es 75 % mayor mayor que x. x
+
x
−
Problema Problema 73. Una pesa no est´ a funcionando correctamente. Si algo es
m´as as ligero que 1000 g, la pesa muestra el peso correcto. Sin embargo, si algo es m´as as pesado o igual que 1000 g, la pesa puede mostrar cualquier n´umero umero por encima de 1000 g. Tenemos 5 pesos A, B , C , D, E cada uno bajo los 1000 g. Cuando se pesan de dos en dos la pesa muestra lo sigui
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66
Restamos (2)
− (1) y obtenemos que C − E = = 100. Por lo tanto, C > E .
Restamos (1)
− (3) y obtenemos que B − A = 100. Por lo tanto, B > A.
Sumamos (4) + (5) y obtenemos que B + C + D + E > 2000. Como B + C = 900, entonces D + E > 1100. Por (5) C + E > 1000, restando estas dos desigualdades se tiene que D C > 100. Por lo tanto D > C .
−
Sumamos (4) + (5) y obtenemos que B + C + D + E > 2000. Como B + E = 800, entonces D + C > 1200. Por (4) B + D > 1000, restando estas dos desigualdades se tiene que C B > 200. Por lo tanto C > B .
−
Finalmente D es el que m´as as pesa. A ndr´es es escrib escr ibee todos to dos los d´ıgitos ıgi tos del Problema 74. Andr´ 1 al 9 en las celdas de una tabla de 3 3, de ma´ ya ha nera que cada celda contiene contiene un d´ıgito. El escrito el 1, 2, 3 y 4, como se muestra en la figura. Dos n´ umeros son considerados como vecinos si sus umeros celdas comparten un borde. Una vez introducidos todos los n´ umeros se da cuenta de que la suma de los umeros vecinos de 9 es 15. ¿Cu´al al es la suma de los vecinos
×
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SOLUCIONES DE LOS PROBLEMAS
67
Concluimos Concluimos que 9 debe estar entre entre 3 y 4, lo que obliga que 8 est´e en el centro, siendo 15 la suma de los vecinos de este.
cuadril´ l´ atero atero Proble Problema ma 75. El cuadri a´ngulos rectos en ABCD tiene angulos los lo s v´erti ertice cess A y D. Los n´umeumeros muestran las areas a´reas de dos de los tri´angulos. angulos. ¿Cu´al al es el area a´rea de ABCD? Soluci´on on Sea E la la intersecci´ on de las diagonales. Notemos que los tri´angulos on angulos ABD y AB C tienen tienen igual ´area, area, pues tienen la misma base AB y la misma altura
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68
la misma altura, proyectada desde C a la prolongaci´on on del segmento DB , como sus bases B E y E D est´an an en la razon 1 : 2, y comparten la altura, la B E C raz´on on entre las areas a´reas tambi´en en es 1 : 2. Dado Da do que el area a´rea del tri´angulo angulo BE E DC es 20. Finalmente, el ´ es 10, el area a´rea del tri´ angulo angulo EDC area area del cuadril´ cuadril´ atero atero es 45 u2. Mar´ıa compiten compiten en la resoluci´ resoluci´ on on de problemas. A Problema Problema 76. Liz y Mar´ cada una de ellas se les da la misma lista de 100 problemas. La primera en resolver cualquiera de estos problemas obtiene 4 puntos, mientras que la segunda en resolverlo obtiene 1 punto. Liz resolvi´o 60 problemas, proble mas, y Mar´ Mar´ıa tambi´ ta mbi´en en resol re solvi vi´ o´ 60 problemas. Juntas, consiguieron 312 puntos. ¿Cu´antos antos problemas fueron resueltos por ambas? Soluci´on on Sea t la cantidad de problemas que resuel-
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SOLUCIONES DE LOS PROBLEMAS
240
−
69
4t + 4w + 5 x = 312 4x + 240 4x + 5x = 312 3x = 168 x = 56
−
−
−
Finalmente, 56 problemas fueron resueltos por Liz y Mar´ Mar´ıa a la vez. ´ Problema 77. David viaja en su bicicleta desde Temuco a su parcela. El deb´ deb´ıa llegar a las 15 : 00, pero gast´ o 32 del tiempo planeado cubriendo 43 de la distancia. Despu´es es de eso, pedale´ o m´ as as lentamente y lleg´o justo a tiempo. ¿Cu´al al es la raz´on on entre la velocidad de la primera parte del viaje y la velocidad de la segunda parte del viaje? Soluci´on on
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70
Luego: 9d 36 3 v1 = 8t = = 3d v2 24 2 4t Finalmente, la raz´on on entre las velocidades de la primera primera y segunda segunda parte del viaje es 3 : 2. Problema Problema 78. En grupo de 25 personas formado por caballeros, ni˜ nos nos
y damas, los caballeros siempre dicen la verdad, los ni˜nos nos siempre mienten, y de las damas algunas mienten y otras dicen la verdad. Cuando se les pregunt´o: o: “¿Es usted una dama?”, dama?”, 12 de ellos dijeron: dijeron: “S “S´´ı”. Cuand Cuandoo se les pregunt´ o: o: “¿Es usted un ni˜no?”, no?”, 8 de ellos dijeron: “S “S´´ı”. ¿Cu´ antos antos caballeros hay en el grupo? Soluci´on on
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SOLUCIONES DE LOS PROBLEMAS
71
Soluci´on on Como en la pizarra tiene que haber al menos 13 n´ umeros umeros divisibles por 13, y exactamente 2 n´ umeros umeros pares, estos tambi´ tambi´en en deben ser m´ ultiplos ultiplos de 13, para que as´ as´ı el m´ as as grande de estos n´ umeros sea el menor valor umeros posible. De este modo elegimos los primeros 11 m´ ultiplos ultiplos de 13 impares y 2 m´ ultiplos de 13 pares menores a 13 21. Finalmente, el menor valor ultiplos posible de M es 13 21 = 273.
·
·
Problema 80. En un estanque hay 16 ho-
jas de lirio de agua formando un patr´ on on de 4 por 4 como se muestra en la imagen. Una rana se sienta en una hoja en una de las esquinas. A continuaci´on, on, salta de una hoja a otra, ya sea horizontal o verticalmente. La rana siempre salta por encima de al menos
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72
Problema Problema 81. Una plaza de 5
× 5 est´a he-
cha de 25 azulejos de 1 1, todos los azulejos con el mismo patr´on, on, tal como el azulejo que se muestra en la figura. Se sabe que en la plaza siempre dos baldosas adyacentes tienen el mismo color a lo largo del borde compartido. El per´ per´ımetro de la plaza se compone de segmentos blancos (lados de tri´ angulos angulos blancos) y grises (lados de tri´angulos angulos grises) de longitud 1. ¿Cu´al a l es el menor n´ umero umero posible de tales segmentos grises de longitud 1?
×
Soluci´on on
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SOLUCIONES DE LOS PROBLEMAS
73
Luego debemos modificar el arreglo inicial e intentar cubrir la plaza dejando 5 segmentos grises alrededor de la plaza, de modo que el interior de la plaza (el cuadrado de 3 3), se pueda cubrir con 4 piezas de la forma
×
, y con una piez ieza de la form orma
.
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74
a˜no n o a a˜ no la suma de sus edades se incrementa en un m´ultiplo no ultiplo de 3, de modo que el incremento de la suma de sus edades debe ser divisible por 3. Teniendo en cuenta que los n´ umeros umeros con 2 d´ıgitos iguales, mayores mayores que 44, son 55, 66, 77, 88 y 99, el unico u ´ nico n´ umero que cumple la condici´ umero on on es 77, pues 77 44 = 33 = 3 11.
−
·
Problema 84. Si ab =
Soluci´on on
1 ¿Cu´al al es el valor de a−3b?. 2
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SOLUCIONES DE LOS PROBLEMAS
75
22014 Problema 86. Calcule el valor de 2013 2
2013
−2 −2
2012
Soluci´on on
22014 22013
−2 −2
2013 2012
22013 (2 = 2012 2 (2 22013
· − 1) · − 1)
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76
Soluci´on on
22 5
55 2
110
2 · 5 = 2
110
·5
= 10110
Notemos que 100 tiene 1 cifra, 101 tiene 2 cifras, 102 tiene 3 cifras, 10n tiene n + 1 cifras. Por lo tanto 10110 tiene 111 cifras. Problema 89. Hector tiene una cuenta de correo electr´ onico onico secreto que
s´olo olo cuatro de sus amigos conocen. Hoy recibi´ o 8 emails a esa cuenta. ¿Cu´ al al
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SOLUCIONES DE LOS PROBLEMAS
77
Soluci´on on Sea r el radio de los cilindros peque˜nos n os y h su altura, al cortar los
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78
no de una caja rectangular es a b c, con a < no Problema Problema 92. El tama˜ b < c. Si aumenta a o b o c en un n´ umero positivo dado, el volumen de la umero
× ×
caja ca ja tambi´en en aumenta. ¿En cu´ al de los siguientes casos el volumen de la al
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SOLUCIONES DE LOS PROBLEMAS
79
Como C tiene 4 puntos, , necesariamente gan´o 1, empat´o 1 y perdi´o 1. Tenemos que en total A + B + C ganaron 4 partidos, por lo que deben
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80
Como queremos que los sumandos sean lo m´ as grande posible, los deas nominadores de cada fracci´ on deben ser lo menor posible. Para a = 2 y on b = 3, se tiene que:
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SOLUCIONES DE LOS PROBLEMAS
Luego n = 10.
81
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82
Problema Problema 99. La l´ınea L pasa por
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Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
Trusted by over 1 million members
Try Scribd FREE for 30 days to access over 125 million titles without ads or interruptions! Start Free Trial Cancel Anytime.
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