Homeschool Math Blog

One of my customers mentioned that this particular puzzle corner from Math Mammoth Grade 1-A was too easy for her son: I admit, it's kind of easy. But, like I told her, I'm sure some children enjoy having an easier puzzle corner in between the others because several people have mentioned they are challenging to their children. But, you can make it more challenging very easily: simply add a number to each equation... and possibly also make (some of) the numbers bigger. Check the examples at http://www.mathmammoth.com/lessons/puzzle_corner_too_easy.php :)

Have you ever encountered this well-known puzzle about the missing $1? Three people rent a room at $30. They pay $10 each and go up to the room. The owner realized he charged too much and it was only supposed to be $25. He sends the bell hop up with the $5. Each of the people keeps $1 and they give the bellhop $2 as they can't share it. So now each person has paid $9 for the room (total $27) and the bell hop has $2... where is the other $1??? See the solution to this and find many more puzzles on my new page of favorite challenging puzzles: http://www.homeschoolmath.net/online/favorite_challenging_puzzles.php Have fun! :^)

Here's something for all of us puzzle lovers: logic imbalance problems invented by Paul Salomon (HT Denise ). You need to order the shapes by their 'weight': Which shape is the heaviest? Which is the second heaviest? Picture by Paul Salomon  Think logically - or write down some inequalities and use algebra. Pretty cool. They are simple, yet captivating. A new, creative idea! Paul also recommends you start making your own imbalance puzzles, as a more 'puzzling' exercise.

Here's a puzzle for you to solve. Fred and Frank are two fitness fanatics on a run from A to B. Fred runs half the way and walks the other half. Frank runs for half the time and walks for the other half. They both run and walk at the same speeds. Who finishes first? Puzzle from Fawn Nguyen's puzzles and brainteasers, set 13 Here's one possible way to solve it. Choose some easy numbers for the distance they travel, for the speeds, and for the time in case of Frank.Then just calculate! Let's say they run 10 miles per hour and walk 5 miles per hour. Let's also say the distance from A to B is 20 miles. Fred runs half the way. So, Fred runs 10 miles = 1 h, and he walks 10 miles = 2 h. So he takes a total of 3 hours. Frank runs half the time and walks half the time. Now, we don't know the time, so let it be t. We can write an equation about the total distance: (distance running) + (distance walking) = 20 miles (1/2t) * 10 + (1/2t) * 5 = 20 5t + 2.5...

This is a solution to the triangle problem I posted here . The picture is below; we are to find the area of the triangle. Since AC is tangent to the circle at D and AB is tangent to the circle at E, then the distances AD and AE are equal. That is, AE must be 2. I can't remember nor find a name for this theorem. It has to do with two tangents from the same point (and it is easy to prove using congruent triangles), saying that those distances are congruent. Similarly, CD and CF are congruent -- both are 3. And also by the same reasoning, BE = BF (because BA and BC are tangets to the circle). I will call BE and BF as x, it's easier to manipulate in an equation. Now, we can solve x fairly easily by applying the Pythagorean theorem to the right triangle ABC. (x + 2) 2 + (x + 3) 2 = (2 + 3) 2 x 2 + 4x + 4 + x 2 + 6x + 9 = 25 2x 2 + 10x + 13 = 25 2x 2 + 10x − 12 = 0 x 2 + 5x − 6 = 0 x = (−5 ± √( 25 − 4(−6))) / 2 x = (−5 ± 7) / 2 x = − 6 or x = ...

This is the solution for the triangle problem with equal areas that I posted earlier. We are given that the areas of the three right triangles are equal, that is the area of the triangle DAE = area of the triangle EBF = area of the triangle FCD. We will make an equation based on that fact. For that, I like to use x as my variable, so I denote the longer side of the rectangle with a, the other side with b, the distance AE with x, and the distance BF with y. We are asked the ratio AE:EB, which is the same as x : (a − x) using my notation, and the ratio BF:FC, which is the same as y : (b − y) using my notation. The area of triangle ADE is its base times altitude divided by 2, or bx/2. The area of triangle EBF is its base times altitude divided by 2, or y(a − x)/2. The area of triangle CDF is its base times altitude divided by 2, or a(b − y)/2. And these three are equal. Basically you just make two equations from the above information, and manipulate your equations until y...

I hope Pat doesn't mind that I copied the image from his blog... He posted this triangle puzzle on his blog and I thought you might enjoy it, too! Basically, we have a triangle DFE inside a rectangle, dividing the rectangle into various triangles.  And, the three areas of fainter color are equal . That is, the area of the triangle DAE = area of the triangle EBF = area of the triangle FCD. (Notice the image is not drawn to scale at all.) And, we're asked to solve the RATIOS AE : EB and BF : FC. The solution is here .

This little puzzle is very intriguing... the answer is simple, SO simple that they claim more educated/intelligent folks have hard time figuring the answer. So go try...! Petals Around the Rose