Archive+Puzzles+61+to+80

Question 80 - Gerry Suppose that four circular disks of equal radius are arranged as in the diagram so that they just touch without overlapping. Suppose that we can just barely fit a circle of radius 1 into the gap between these four circles. What is the radius of one of these four circles? Remember to give an exact answer -- no approximations allowed! Solution 80 Pythagorean theorem- a squared +b squared=c squared 1+x squared times 2 = 4x squared 2+4x+2x squared=4 x squared 2+4x=2x squared Can someone who knows the quadratic formula continue?~GArett -2x^2 + 4x +2 =0 The quadratic formula is ax^2+bx+c=0 x=-b+(sqrt root of b^2-4ac)/2a OR -b - (sqrt root of b^2-4ac)/2a Here, x=-4+(sqrt root of 16+16)/-4 or -4-(sqrt root of 32)/-4 Working positive x=-4+sqrt of 32 / -4

Working negative x=-4-sqrt of 32 / -4

Although the numerator of the 'negative' is negative, it also has a negative denominator. Hence, x is (4+ sqrt of 32)/4~Alex Or if you want, it's 1+sqrt of 2. Even more accurate, it's **2.4142135.** btw, i know this is considered cheating, but simply right' clicking on the picture then selecting 'properties would show you the URL he got it from, for example, this one was from [] ~Garett I prefer doing things legitimately.~Alex ... I don't cheat, i used that to check my answers btw, a more accurate answer would be: 2.4142135623730950488016887242097~Garett

Question 79 - Gerry Two spiders walk across the //xy//-plane. The //x// and //y// coordinates of the first spider at time //t// seconds are //x = 10 + t// and //y = t2 - 4 t + 17// The //x// and //y// coordinates of the second spider at time //t// seconds are

//x = 2t// and //y = 4 t - 12// The paths of these two spiders touch at one point -- find that point! Be sure to explain your reasoning along with giving your answer. //Hint:// The two spiders do not get to that point at the same time!

Solution 79

I so do not understand the question~Garett Urgh... this calls for a table. (SPIDER 1)Ignore table at top of pg(if u see 1), that was on accident. My bad.:(CONCLUSION: I dont know~Alex (SPIDER 2)
 * t || 1 || 2 || 3 || 4 || 5 || 6 || 7 || 8 || 9 || 10 || 11 || 12 || 13 || 14 || 15 || 16 ||
 * x || 11 || 12 || 13 || 14 || 15 || 16 || 17 || 18 || 19 || 20 || 21 || 22 || 23 || 24 || 25 || 26 ||
 * y || 14 || 13 || 14 || 17 || 22 || 29 || 38 || 49 || 62 || 77 || 94 || 113 || 134 || 157 || 182 || 209 ||
 * t || 1 || 2 || 3 || 4 || 5 || 6 || 7 || 8 || 9 || 10 || 11 || 12 || 13 || 14 || 15 || 16 ||
 * x || 2 || 4 || 6 || 8 || 10 || 12 || 14 || 16 || 18 || 20 || 22 || 24 || 26 || 28 || 30 || 32 ||
 * y || -8 || -4 || 0 || 4 || 8 || 12 || 16 || 20 || 24 || 28 || 32 || 36 || 40 || 42 || 44 || 46 ||

Question 78 - Gerry The figure at right shows an infinite pattern of nested circles and squares. Suppose that the red area is exactly 1. What is the length of the largest square's side? You may give an approximation, but it must be accurate to within 1/100. Good luck!

Solution 78 0.43~Garett 0.79~Alex(BTW, the circle's more than half of the square.) Alex, you are closer but the answer is actually about **1.5263997**. OMGWTF? How do you prove it?~Alex maybe by ratio its like if there was only the biggest square and circle, the red area would be 2r squared - pi r squared the ratio of the red area is to the yellow area would be 4-pi : pi but if we insert the second square, the yellow area would be pi r squared- 2 r squared so the new ratio of the red and yellow areas, after inserting all the other squares and circles 4-pi:pi-2 now the problem is : WHAT IS PI? if pi is 22 over 7,the ratio is: 6 over 7 : 8 over 7 1: 1 and 1 third If it is 3.14, the ratio would be 0.86 : 1.14 but according to the computer's pi 1:1.3298961831627438472393536346285 psps for my last answer now the answer: 1.5263997455328482356223881550839~Garett

Question 77 - Gerry While flying over farmland, a pilot notices the rectangular shape of the fields below. She sketches the lines that divide the fields. When she returns to the airport, she wonders how many different rectangles can be formed by the lines drawn? HINT: Don't forget that a square is also a rectangle. Solution 77 1+2+2+4+3+6=18 rectangles ~Daniel Correct
 * [[image:http://mathforum.org/k12/k12puzzles/critical.thinking/farmland.gif align="center" caption="Puzzle Picture"]] ||
 * Puzzle Picture ||

Question 76 - Gerry To move their armies, the Romans built over 50,000 miles of roads. Imagine driving all those miles! Now imagine driving those miles in the first gasoline-driven car that has only three wheels and could reach a top speed of about 10 miles per hour. For safety's sake, let's bring along a spare tire. As you drive the 50,000 miles, you rotate the spare with the other tires so that all four tires get the same amount of wear. Can you figure out how many miles of wear each tire accumulates?

Solution 76 50000/4=12500 12500/3=4166 2/3 The car drives for 4166 2/3 miles Tyre 1: 4166 2/3 Tyre 2: 4166 2/3 Tyre 3: 4166 2/3 Tyre 4: 0(SPARE) Switch out Tyre 1 for 4, drive another 4166 2/3 miles Tyre 1: 4166 2/3(SPARE) Tyre 2: 8333 1/3 Tyre 3: 8333 1/3 Tyre 4: 4166 2/3 Switch Tyre 2 for 1, drive another 4166 2/3 miles Tyre 1: 8333 1/3 Tyre 2: 8333 1/3(SPARE) Tyre 3: 12500 Tyre 4: 8333 1/3 Switch Tyre 3 for 2, drive another 4166 2/3 miles Tyre 1: 12500 Tyre 2: 12500 Tyre 3: 12500(SPARE) Tyre 4: 12500 Switch Tyre 4 for 3. Repeat for another 2 times, and you have your 50000 miles driven, with equal amount of tear.~Alex Yep.

Question 75 - Gerry

It's amazing how many good puzzles come from just simple old squares. Suppose we start with a square //ABCD// with sides of length 1 and cut off two opposite corners (such as the purple triangles in the figure at right) so that the hexagon //AEFCGH// that remains has all sides of equal length. Your challenge is to find the area of the rectangle //EFGH// (shown in yellow in the figure) Solution 75

(sqrt of 2) cubed over 3 + (sqrt of 2) cubed~Garett Correct

Question 74 - Gerry Ancient Egyptian pyramids were built as royal tombs. Within these massive stone structures were rooms, halls, and connecting passageways. Look at the figure below. Can you draw four paths that connect the matching symbols? The paths may not cross, they may not enter a non-matching pyramid, nor may they go outside the larger pyramid boundary.
 * [[image:http://mathforum.org/k12/k12puzzles/critical.thinking/pyramid.gif width="246" height="208" align="center" caption="Puzzle Picture"]] ||
 * Puzzle Picture ||

Solution 74

it is impossible. looka t it. the eye pictures, no matter how you connect them, wil block the lines or the birds. so, it's impossible. it is possible~gerry invalid, kai xuan is correct, becuz the top eye has to bend to reach the bottom, and that will block either the lines or birds, and if you do the bird and line first it forms a 'cage' that prevents the eye to enter... therefore this question can only be described by a 10 letter word: IMPOSSIBLE!~alex It is possible! Look! ~Gerry

Question 73 - Gerry Can you construct nine triangles by drawing three straight lines through a capital **M**? Solution 73

No. Duh~Alex Yes! ~Gerry

Question 72 - Gerry The red square ABCD at right has side length 50. The point M is the midpoint of side AB, and the green circle passes through C, D, and M. What is the radius of that circle? Solution 72 31.25~Garett correct

Question 71 - Gerry Can you connect 9 dots laid out 3x3 using 4 straight lines, without lifting your pencil from the paper? Solution 71 Umm..you have to think out of the box?~Jun Yuan

Step 1: Draw a line on the top row of dots, but draw the line a little bit to the outside. Step 2: Draw the 2nd line diagonally left downward, drawing the line a little bit to the otuside too. Step 3: Draw the 3rd line straight up. Step 4: Draw the 4th line diagonally right downward. If you don't understand, I can show you guys in school. ~ Alton Solved ~Gerry

Question 70 - Gerry Use the digits in the years 2009, 2007, 2006, 2005, 2004, 2003, 2002, 2001, 2000, 1999, or 1998 and the operations of +, -, x, ÷, sqrt (square root), ^ (raise to a power), and ! (factorial), along with grouping symbols, to write expressions for the counting numbers 1 through 100. I don' t expect anyone to do this question.~Gerry (P.s. I don't know the answers myself)

Solution 70 1=2-0-0-1 2=2-0-0-0 3=20x0+3 4=20x0+4 5=20x0+5 6=20x0+6 7=20x0+7 8=2+0-0+6 9=20x0+9 10=1+9-9+9 11=2+0+0+9 12=(2+0+0)x6 13=20-0-7 14=(2-0-0)x7 15=20-0-5 16=20-0-4 17=20-0-3 18=20-0-2 19=20-0-1 20=20-0-0 21=20+0+1 22=20+0+2 23=20+0+3 24=20+0+4 25=20+0+5 26=20+0+6 27=20+0+7 28=1+9+9+9 29=20+0+9 30=20+0!+9 31=2^5-0!-0(seriously, i dunno if this is legal) 32=2^5-0-0(see 31) 33=2^5+0!-0(see 32) 34=2^5+0!+0!(see 33) 35=6^2-0!+0(see 34) 36=(2+0!+0!)x9 37=6^2+0!+0)(see 35) 38=6^2+0!+0!(see 36) 39=(dunno) 40=(20+0)x2 41=20x2+0! 42=42+0+0 43=42+0!+0 44=42+0!+0! (Is stopping for now...)~Alex On a side note, literally all the numbers can be given the solution 0/0... LOLZ at a (42x0)/0, simply because 39x0=0, 44x0=0 etc.(yup, thats illegal, btw ,gerry, i really need your confirmation whether we can swap numbers about...)

Question 69 - Gerry
 * A logic problem: What question should the princess ask?**

A princess visits an island inhabited by two tribes. Members of one tribe always tell the truth, and members of the other tribe always lie. The princess comes to a fork in the road. She needs to know which road leads to the castle so as to avoid the fire-breathing dragon and rescue the prince from the wizard holding him captive in the castle. (Although the princess doesn't know it, the south road leads to the castle and the north road leads to the dragon.) Standing at this fork in the road is a member of each tribe, but the princess can't tell which tribe each belongs to. What question should she ask to find the road to the castle?

Solution 69 The princess should ask: If I were to ask the member of the other tribe where the castle was, where would he point to? ~Garett (Met a similar question b4) Correct~ Gerry It was in Fiendish's Bk of brainbenders, where u r supposed to escape from Urgum's castle.~Alex

Question 68 - Gerry
 * Suppose you flip a coin and bet that it will come up tails. Since you are equally likely to get heads or tails, the probability of tails is 50%. This means that if you try this bet often, you should win about half the time.


 * What if somebody offered to bet that at least two people in your math class had the same birthday? Would you take the bet?** || [[image:http://mathforum.org/dr.math/faq/images/balloons2.gif width="136" height="194" caption="external image balloons2.gif"]] ||

This question is more complicated than flipping a coin, because the chance of finding two people with the same birthday depends on the number of people you ask. If there were only one other person in your math class, you might be surprised to find out that she had the same birthday as you. If there were a pair of people with the same birthday in a class of 366 people, would you still be surprised?
 * How large must a class be to make the probability of finding two people with the same birthday at least 50%?**

Solution 68 srsly, stop reading 'Do You Feel Lucky?' Answer is 23. You arent the only one that has read the book! correct, but how about the first question of course it's a no. lolz~Alex

Question 67 - Gerry Imagine that the set of Monty Hall's game show //Let's Make a Deal// has three closed doors. Behind one of these doors is a car; behind the other two are goats. The contestant does not know where the car is, but Monty Hall does. The contestant picks a door and Monty opens one of the remaining doors, one he knows //doesn't// hide the car. If the contestant has already chosen the correct door, Monty is equally likely to open either of the two remaining doors. After Monty has shown a goat behind the door that he opens, the contestant is always given the option to switch doors. **What is the probability of winning the car if she stays with her first choice? What if she decides to switch?** Solution 67 LOL MM. Without changing, the guy has 1/3 chance of winning rite?
 * Let's Make a Deal!**

Well, if he decides to switch:

Win, then change:1/6 possiblity, contestant loses Lose, then change:1/3 possiblity, contestant wins. So, 1/3+1/3=2/3, so if he switches, contestant has a 2/3 chance of winning. Correct~ Gerry

Question 66 - Gerry You must draw a line from each house to each utility, without the lines ever crossing. **Can you connect the houses to the utilities?** Solution 66
 * There are three houses and three utilities:**

Can the lines be curved? If they can, I can show it to you in school. ~ Alton Yes, they can. But they must be on a piece of paper and not a 3-d surface Solved~Gerry

Question 65 - Gerry Solution 65
 * A donkey is attached by a rope to a point on the perimeter of a circular field. How long should the rope be in terms of the radius of the field so that the donkey can reach exactly half the field and eat half the grass?**

Uh... Im afraid i dont understand this question...

Question 64 - Gerry Karen can row a boat 10 kilometers per hour in still water. In a river where the current is 5 kilometers per hour, it takes her 4 hours longer to row a given distance upstream than to travel the same distance downstream. Find how long it takes her to row upstream, how long to row downstream, and how many kilometers she rows. Solution 64 On upstream, Karen's spd is 5 km/h, downstream 15 km/h. Assuming distance is 60 km, upstream is 12, downstream is 4. Assuming distance is 120 km, upstream is 24, downstream is 8. Assuming distance is 30 km, upstream is 6, downstream is 2. Therefore she takes 6 hours upstream, 2 hrs downstream, distance is 30 km~Alex Correct~Gerry

Question 63 - Alex What is the possiblity that you will win 4-D, Toto, The Big Sweep all in ONE WEEK? Solution 63

Question 62 - Alex Lazy Leslie told her superior one day:"There's no way will I have any time to work. There are 365 days in a year. I sleep 12 hours every day, so I just wasted half a year, just sleepin'. 365/2=182.5, and thus, its 182.5 days left to do work. I have 11 public holidays, work on alternate Saturdays, do not work on Sundays, and have a 24-day paid leave, so that's 11+26+52+24=113 days, so I have 69.5 days left to do work. I wake up at 7, go to work at 8, leave work at 6, and sleep by 7. 2 hours a working day=139 hours wasted. That's 5 days and 19 hours. Subtract from 69 days 12 hours... (SUBTRACTION SUM BELOW) 69 days 12 hours-5 days 19 hours=68 days 36 hours-5 days 19 hours=63 days 17 hours. (END OF SUBTRACTION SUM) Yep, that's 63 days 17 hours left for work. I have a lunch break of 1 hour every day, so it's 69.5 hours down the drain. Subtract from 63 days 17 hours.. (ANOTHER SUBTRACTION SUM) 63 days 17 hours-69.5 hours=60 days 89 hours-69.5 hours=60 days 19.5 hours. (EOSS) So yeah, I have 60 days 19.5 hours to do work. Now, tell me sir, how the heck am I gonna do work in just enough time to LAST 2 and a half days? Surely this is commiting suicide! To think I have toilet breaks in there too..." Unfortunately, her superior was a smart fellow, and instantly saw through the deliberate errors Leslie put in her testimony. Leslie was punished with a pay cut, lunch break removal, toilet break removal, and lengthened working hours. What are the lies in Leslie's testimony? How long should she have for work originally? Solution 62

Well, to start everything off, I would like to say good afternoon. Then, I shall start my speech with a "duh". Then another "duh". And another. Now, I will start. First of ALL, she should not be starting off with "12h /day", 'cause this make her days at work 69.5 days after sleeping + free time (which is (1/2 x 365)h +(11+26+52+24) while it should have been (365 - 113)/2, which is 126. SO, I hope that my answer has AT LEAST 1 point that is right. Thank you.
 * Second of ALL, why did she count sleeping time AGAIN(2x)? (Should be 126)
 * Third of ALL, back to the first point. It should be 126h, not 69.5h. So, it becomes 120.75 days, which is 120 days and 18 hours.***

~Keith
 * Sorry. This should be "Secondly".
 * Sorry for not really being sorry. Sorry. This should be "Lastly".
 * In case you were wondering what this is, it has no meaning. Haha! I'm not sorry at all!

Question 61 - Alton I deleted Question 60 due to cheaters. This time the answer will not be on recent changes anymore, you'll have to use your brain, Garett.

ABCDE + FGHIJ = DFHIBJ What is A, B, C, D, E, F, G, H, I and J?

Solution 61 So, which brain, left or right? Who cares, now let's start solving that question E=0 A and F have to be more than 5 D+I =B+10 C+H+10=I+10/20~Garett. I continue whe I have the chance. Stop cheating to solve maths, thats not learning. BTW, E=0. Since its a (insert random 5-digit number)+(insert another random 5-digit number here)=(Insert 6-digit sum here), A cannot be 1, since 9+9+1=19, thus A does not equal one. On a side note, D=1.~Alex (Ugh, this is exploding my brain off.) Continuing G's notes, 1+I =B+10 Therefore B=I-9???(How the screw is this even possible... E=0...)~Alex (Brain overheating) Anyway, I will assume something's wrong with G, so lets get on with this thing. So far, established A:?(Cannot equal 1) B:? C:? D:1 E:0 F:? G:? H:? I:? J:? Therefore, ABC10+FGHIJ=1FHIBJ But wait... something is seriously wrong here. If A+F=1F(-1), this is seriously insane. If A is 2, F must be 7 to 9(so that the D can exist) but then the result is far from 17...(potential solutions bolded) a=3, f=6 to 9(NO, Result not near 16)9, 10, 11, 12 a=4, f=5 to 9(NO, result not near 15)9, 10 ,11 ,12, 13 a=5, f=4 to 9(not possible)9, 11,12,13,14 a=6, f=3 to 9(n/p)9, 10,11,13,14,15 a=7, f=2 to 9(n/p)9, 10, 11,12,13,15,16 a=8, f=2 to 9(n/p)10,11,12,13,14,15,17 a=9, f=2 to 8(possible)**11,12,13,14,15,16,17** This proves that a has to be 9. Established: A=9 This means B+G is in fact, more than or equal to 10.(Tracking:9BC10+FGHIJ=1FHIBJ) WOOT:)~Alex(is stopping for now!:)) Lets see... leads: J is the least in priority.(duh) C+H=I, since we know that 1 is already used, 1+I=B Our carryover should occur somewhere around B+G. Finding F is quite tricky here. We have already used 0,1 and 9. I cannot be 8 in that case. And I also cannot be 2(1+1 or 0+2) So... so far... 1+I=B C+H=I B+G=H+10 Wow, thats a load of simultaneous equations!(Actually, they arent really simultaneous...) 1+C+H=B 1+C+H+G=H+10 1+C+G=10 C+G=9 Since 1 is already used, we know C and G both cannot be 8. Therefore, lets list the possiblities as to what we have so far. A:9 B:3,4,5,6,7,8 C:2,3,4,5,6,7 D:1 E:0 F:Ranging from 2-8 G:2,3,4,5,6,7 H:2,3,4,5,6,7 I:2,3,4,5,6,7 J:Any remaining number H=I-C B=I+1 B-H=1+C So we know that the diff between B and H is C plus 1. And look what do we have here?! B cannot be 3 no longer, because... H=2, 1=1+C, C=0(No) H=4, -1=1+C, C=-2(No, duh) H=5, -2=1+C, C=-3(Seriously obviously no) and so on... so we eliminate 3 for option B.(B:4,5,6,7,8) (is still stopping for now...)~Alex Ok, back to this thing... B cannot equal 3, so lets try 4 H=2, 2=1+C, C=1(No) H=4, 0=1+C, C=-1(No, will stop_ Proved: B cannot equal 3 or 4. Lets try 5. H=2, 3=1+C, C=2(No way) H=3, 2=1+C, C=1(Will stop) Lets try 6. H=2, 4=1+C, **C=3**(Finally, a ray of hope!!!) H=3, 3=1+C, C=2(Woot, another ray of hope!!!) H=4, 2=1+C(Self-explanatory) Ok, now 7 H=2, 5=1+C, C=4 H=3, 4=1+C, C=3(NONONO!!!) H=4, 3=1+C, C=2 H=5, 2=1+C(Already self-explanatory) 8... H=2, 6=1+C, C=5 H=3, 5=1+C, C=4 H=4, 4=1+C, C=3 H=5, 3=1+C, C=2 Therefore... Lets make a note of all the possiblities. B=6 H=2 C=3 B=6 H=3 C=2 B=7 H=2 C=4 B=7 H=4 C=2 B=8 H=2 C=5 B=8 H=3 C=4 B=8 H=4 C=3 B=8 H=5 C=2 Wat!? 8 possiblities! Oh well... H=I-C B=I+1 Tracking here... A:9 B:6,7,8 C:2,3,4,5 D:1 E:0 F:2nd least in priority G:4,5,6,7 H:2,3,4,5 I:5,6,7 J: Least in priority We know that since H=I-C, thus H+C=I(Lol Hwa Chong but who cares) Tracking:9BC10+FGHIJ=1FHIBJ (Is stopping for now, what a load of work!)~Alex ok, back to the question, I will have to do a lot of testing here. Considering that I have already got the possiblities. 1. 96310+FG25J=1F256J From the looks of it, this is quite likely. The problem starts with G. G has to be a 6, so that the 2 can be there. Therefore... VERDICT: Wrong, 6 is repeated. 2. 96210+FG35J=1F356J Now, this may not seem different from before, but this actually is more likely. Start trashing G. We know that 2+3=5 So 6+G=13, results in G=7 96210+F735J=1F356J Numbers used: 0,1,2,3,5,6,7,9 So, that leaves us with 2 answers 96210+47358=143568 96210+87354=183564 And what do you know, BOTH ARE CORRECT! Verdict: Correct, with 2 answers!(Will continue later)~Alex. 3.97410+FG26J=1F267J G=5 97410+F526J=1F267J And thus... 97410+35268=132678 97410+85263=182673 VERDICT: Correct. 2 answers! 4.97210+FG46J=1F467J 7+7=14, and you know what happens next... 5.98510+FG27J=1F278J G=4 So, 98510+F427J=1F278J 98510+34276=132786 98510+64273=162783 Verdict: 2 correct answers! 6.98410+FG37J=1F374J G=5 98410+F537J=1F374J 98410+25376=123746 98410+65372=163742 Verdict: 2 correct Answers!!!!!!!!!!!!! 7.98310+FG47J=1F478J G is 6. so... 98310+F647J=1F478J 98310+26475=124785 98310+56472=154782 Verdict: Almost to the end. 10 correct answers so far. 8.98210+FG57J=1F578J Verdict: Invalid answer because 7 is repeated. So therefore, to wrap up, the 10 possible answers are: 2.96210+87354=183564 3.97410+35268=132678 4.97410+85263=182673 5.98510+34276=132786 6.98510+64273=162783 7.98410+25376=123746 8.98410+65372=163742 9.98310+26475=124785 10.98310+56472=154782** The End.~Alex(who has painstakingly typed out at least 40 sentences just for this question...) P.S. Do i get a record for the longest solution ever? LOL P.P.S. Alton, u forgot that F and J are seperate from the ABCDEGHI confusion, making multiple solutions possible. If you see carefully. you will notice that some solutions are 'repeated' simply by rotating the positions of F and J, which makes 0 difference. Thank you for your time. The end.
 * yawn* lets continue. We know C+G=9. 1 is already used, so no 8s allowed.(Tracking:9BC10+FGHIJ=1FHIBJ)
 * 1.96210+47358=143568