Interesting Math Problems

[Animal]

Two circles of radius 7 units, A and B, are tangent to one another. Another circle of radius 7 units, O, overlaps circles A and B. The diameter of circle O sits on the same line that is tangent to the tops of circles A and B. What is the area of the overlap of circle O over circles A and B?

55.938?

[QillerDaemon]

Two circles of radius 7 units, A and B, are tangent to one another. Another circle of radius 7 units, O, overlaps circles A and B. The diameter of circle O sits on the same line that is tangent to the tops of circles A and B. What is the area of the overlap of circle O over circles A and B?

55.938?

You're right, but you didn't show your work!

[Animal]

Two circles of radius 7 units, A and B, are tangent to one another. Another circle of radius 7 units, O, overlaps circles A and B. The diameter of circle O sits on the same line that is tangent to the tops of circles A and B. What is the area of the overlap of circle O over circles A and B?

55.938?

You're right, but you didn't show your work!

the work was pretty simple.

The shape we are trying to calculate the area of are two identical "eye" shapes that are formed inside of two 7x7 inch squares. So, I will figure the area of one of them and then double it.

The total square to start with is 7x7 = 49

Then we need to subtract out two filet areas created by the radius of 7. The area of one of those filet areas is r^2 * (1 - PI/4) or 49 * (1 - 3.1416/4)

Which is 10.5155. We have two of those that need to be subtracted from 49. 49 - 10.5155 - 10.5155 = 27.969.

Now, we have to double that to get the area inside circle A and B. 27.969 * 2 = 55.9380

[Bluespruce1964]

I though this said interesting Meth problem

[QillerDaemon]

Two circles of radius 7 units, A and B, are tangent to one another. Another circle of radius 7 units, O, overlaps circles A and B. The diameter of circle O sits on the same line that is tangent to the tops of circles A and B. What is the area of the overlap of circle O over circles A and B?

55.938?

You're right, but you didn't show your work!

the work was pretty simple.

The shape we are trying to calculate the area of are two identical "eye" shapes that are formed inside of two 7x7 inch squares. So, I will figure the area of one of them and then double it.

The total square to start with is 7x7 = 49

Then we need to subtract out two filet areas created by the radius of 7. The area of one of those filet areas is r^2 * (1 - PI/4) or 49 * (1 - 3.1416/4)

Which is 10.5155. We have two of those that need to be subtracted from 49. 49 - 10.5155 - 10.5155 = 27.969.

Now, we have to double that to get the area inside circle A and B. 27.969 * 2 = 55.9380

[QillerDaemon]

A circle is drawn, then inside the circle is drawn a semi-circle. The flat side of the semi-circle forms a chord to the circumference of the big circle. The flat side of the semi-circle is 8 units from the circumference of the big circle. The round side of the semi-circle is 48 units from the other side of the big circle. What is the area of the big circle *not* including the semi-circle? (hint: intersecting chords theorem)

[QillerDaemon]

Two semi-circles of unequal size site flat side down on the same surface line. They are separated by some unknown distance. The larger semi-circle is 21 units, the smaller semi-circle is 6 units. The tangent line connecting the round sides of the semi-circles is 36 units. How far apart are the two semi-circles?

[Animal]

Two semi-circles of unequal size site flat side down on the same surface line. They are separated by some unknown distance. The larger semi-circle is 21 units, the smaller semi-circle is 6 units. The tangent line connecting the round sides of the semi-circles is 36 units. How far apart are the two semi-circles?

I just saw these.

when you say the semi circles are 21 units and 6 units. is that their radius or diameter? I am assuming its the diameter.

[Animal]

Two semi-circles of unequal size site flat side down on the same surface line. They are separated by some unknown distance. The larger semi-circle is 21 units, the smaller semi-circle is 6 units. The tangent line connecting the round sides of the semi-circles is 36 units. How far apart are the two semi-circles?

I just saw these.

when you say the semi circles are 21 units and 6 units. is that their radius or diameter? I am assuming its the diameter.

If those numbers are the diameters of the semicircles, then the distance apart is 21.71 units.

If those numbers are the radius of the semicircles, then the distance apart is 5.726 units.

[stonedmegman]

[QillerDaemon]

Two semi-circles of unequal size site flat side down on the same surface line. They are separated by some unknown distance. The larger semi-circle is 21 units, the smaller semi-circle is 6 units. The tangent line connecting the round sides of the semi-circles is 36 units. How far apart are the two semi-circles?

I just saw these.

when you say the semi circles are 21 units and 6 units. is that their radius or diameter? I am assuming its the diameter.

Ooof, you're right. 21 and 6 units in radius, not diameter.

[Homebrew]

[Animal]

Two semi-circles of unequal size site flat side down on the same surface line. They are separated by some unknown distance. The larger semi-circle is 21 units, the smaller semi-circle is 6 units. The tangent line connecting the round sides of the semi-circles is 36 units. How far apart are the two semi-circles?

I just saw these.

when you say the semi circles are 21 units and 6 units. is that their radius or diameter? I am assuming its the diameter.

If those numbers are the diameters of the semicircles, then the distance apart is 21.71 units.

If those numbers are the radius of the semicircles, then the distance apart is 5.726 units.

ahem. BUMP

[QillerDaemon]

If those numbers are the radius of the semicircles, then the distance apart is 5.726 units.

Nope.

[Animal]

If those numbers are the radius of the semicircles, then the distance apart is 5.726 units.

Nope.

Two semi-circles of unequal size site flat side down on the same surface line. They are separated by some unknown distance. The larger semi-circle is 21 units, the smaller semi-circle is 6 units. The tangent line connecting the round sides of the semi-circles is 36 units. How far apart are the two semi-circles?

I just saw these.

when you say the semi circles are 21 units and 6 units. is that their radius or diameter? I am assuming its the diameter.

If those numbers are the diameters of the semicircles, then the distance apart is 21.71 units.

If those numbers are the radius of the semicircles, then the distance apart is 5.726 units.

Well, fuck. Let me go back through this.

Two Semicircles. R=21. and R=6. They are some distance apart lying flat on same plane. The tangent line between the two of them is 36. (Oh, I know what I did wrong.)

So, if the tangent is 36. Then they are exactly 12 units apart.

Its kind of confusing on the work. The tangent line, if you carry it all the way to where it intersects the flat plane the semicircles rest on......

The distance of that tangent line from the smaller semicircle to the intersection is X. So that line is 36+x from the intersection to the large circle. And it is X from the intersection to the small circle. The theta angle where it meets the intersection would be the same for both right triangles created where it meets each semicircle. The short leg of each triangle being their radius.


So, tan Theta = (36+x)/21 which also equals x/6. If you solve for X (the tangent line to the small circle) you get 14.4 units. Solving for theta with tan Theta = 14.4/6 we get Theta of 22.619865 degrees. If you use Sin Theta to figure the length of the flat line to the radius point it would be 15.6 units.

Now, the length of the tangent to the large circle is 36+x = 36 + 14.4 = 50.4 units. Using sin Theta to figure the length of the flat line to the radius of the large semicircle we get 54.6. Now, the distance between the two circles would be 54.6 - 21 - 6 - 15.6 = 12.

[QillerDaemon]

Now you win the prize! Yes, the distance between them is 12 units.

You solved it with trig, and there's nothing wrong with that. I solved it with geometry.

The tangent line of 36 can make a rectangle with the 6 unit radius of the smaller semi-circle. Under that rectangle is a right triangle with a long leg of 36, a short leg of (21 - 6 =) 15, and hypotenuse of (21 + x + 6 =) x + 27. Using the Pythagorean Theorem:

36² + 15² = (x + 27)² and arranging and reducing this becomes x² + 54x - 792 = 0.
This factors into (x + 66) and (x - 12), the first gives us a negative length, the second gives us the correct answer.

Putting that back into the formula of that right triangle produces a Pythagorean triple -> 3x (5, 12, 13)!

[Animal]

Now you win the prize! Yes, the distance between them is 12 units.

You solved it with trig, and there's nothing wrong with that. I solved it with geometry.

The tangent line of 36 can make a rectangle with the 6 unit radius of the smaller semi-circle. Under that rectangle is a right triangle with a long leg of 36, a short leg of (21 - 6 =) 15, and hypotenuse of (21 + x + 6 =) x + 27. Using the Pythagorean Theorem:

36² + 15² = (x + 27)² and arranging and reducing this becomes x² + 54x - 792 = 0.
This factors into (x + 66) and (x - 12), the first gives us a negative length, the second gives us the correct answer.

Putting that back into the formula of that right triangle produces a Pythagorean triple -> 3x (5, 12, 13)!

and that is actually a much cleaner solution. I like it. i'm not a big fan of using trig since you have to have a calculator with those functions. but that's the road it started down....

[QillerDaemon]

A card manufacturer makes a set of cards that have a letter on one side of a card and a number on the other side.

One rule with the set of these cards is that any card with the letter D on one side has the number three on the other side.

As a test, four cards are set out on the table: a card with the letter D, a card with the letter K, a card with the number 3, and a card with the number 7.

Which cards have to be turned over to check whether the manufacturer has printed the set of cards correctly? (Turning over all four is not correct!)

[stonedmegman]

A card manufacturer makes a set of cards that have a letter on one side of a card and a number on the other side.

One rule with the set of these cards is that any card with the letter D on one side has the number three on the other side.

As a test, four cards are set out on the table: a card with the letter D, a card with the letter K, a card with the number 3, and a card with the number 7.

Which cards have to be turned over to check whether the manufacturer has printed the set of cards correctly? (Turning over all four is not correct!)

1

[Animal]

A card manufacturer makes a set of cards that have a letter on one side of a card and a number on the other side.

One rule with the set of these cards is that any card with the letter D on one side has the number three on the other side.

As a test, four cards are set out on the table: a card with the letter D, a card with the letter K, a card with the number 3, and a card with the number 7.

Which cards have to be turned over to check whether the manufacturer has printed the set of cards correctly? (Turning over all four is not correct!)

Well, based on the rule, the only thing that matters are cards that have a D or a 3 on them, so we can rule out worrying anything about the "K" or the "7". So, now we are left with a card that has a "D" and a card that has a "3". Are they printed correctly? First, let's think about the "3". Does it matter what is on the other side? No. Because if it was a "K" on the other side, then there is no rule that a K can't have a 3 on the back side.

So, we only have to turn over the "D" to make sure they followed the rule and put a "3" on the other side.

[Animal]

A card manufacturer makes a set of cards that have a letter on one side of a card and a number on the other side.

One rule with the set of these cards is that any card with the letter D on one side has the number three on the other side.

As a test, four cards are set out on the table: a card with the letter D, a card with the letter K, a card with the number 3, and a card with the number 7.

Which cards have to be turned over to check whether the manufacturer has printed the set of cards correctly? (Turning over all four is not correct!)

Well, based on the rule, the only thing that matters are cards that have a D or a 3 on them, so we can rule out worrying anything about the "K" or the "7". So, now we are left with a card that has a "D" and a card that has a "3". Are they printed correctly? First, let's think about the "3". Does it matter what is on the other side? No. Because if it was a "K" on the other side, then there is no rule that a K can't have a 3 on the back side.

So, we only have to turn over the "D" to make sure they followed the rule and put a "3" on the other side.

Thinking about this some more, I think you would have to turn over (3 cards) every single card except the 3.

1. You have to turn over the "D" to make sure they followed the rule and put a 3 on the back.
2. You have to turn over the "K" to make sure that there is not a "D" on the other side, which would make it printed wrong.
3. You have to turn over the "7" to make sure that there is not a "D" on the other side.

But, it doesn't matter what is printed on the other side of the "3". If its a "D" then its printed right. If its any other letter, then it doesn't matter because there is no rule about that.

[QillerDaemon]

Thinking about this some more, I think you would have to turn over (3 cards) every single card except the 3.

1. You have to turn over the "D" to make sure they followed the rule and put a 3 on the back.
2. You have to turn over the "K" to make sure that there is not a "D" on the other side, which would make it printed wrong.
3. You have to turn over the "7" to make sure that there is not a "D" on the other side.

But, it doesn't matter what is printed on the other side of the "3". If its a "D" then its printed right. If its any other letter, then it doesn't matter because there is no rule about that.

Close, but still not sparking that bud...

The rule says if a card has a D on one side, then it has a three on the other. So by logic, "if D, then 3." The converse statement, "if 3, then D" is not always true, if the positive is true. "If it is raining outside, you need an umbrella" may be true, but its converse "if you need an umbrella, it's raining outside" is not, as you might need an umbrella for other reasons. The inverse of "if D, then 3" is "if not D, then not 3" is also not always true. "If it is not training, you don't need an umbrella" is not true. On the contrapositive "if not 3, then not D" is true if "if D, then 3" is true. "If you don't need an umbrella, then it is not raining" is a true statement if the original statement is true.

So for "if D, then 3", the converse is "if 3, then D", the inverse is "if not D, then not 3", and the contrapositive is "if not 3, then not D".
That is by the statement of the original rule "if a card has a D on one face, it has a three on the other."
The converse says "if a card has a three on one face, it has a D on the other." -- not necessarily true.
The inverse says "if a card does not have a D on one face, then it does not have a 3 on the other." -- also not necessarily true.
The contrapositive says "if a card does not have a 3 on one face, it does not have a D on the other." -- this is necessarily true.

The cards are chosen to illustrate each kind of statement:

The D face is "if D, then 3" case. If we turn over the D and the other side is 3, the rule is satisfied. If the other side is not a three, the rule is broken and we have a bad set of cards.

The K face is the "if not D, then not 3" case. We turn over the K card and can find either a 3 or some other number, neither case breaks the original rule. So we don't need to turn over the K card.

The 3 face is the "if 3, then D" case. As above, we turn over the 3 card and can find either a D or some other letter, and again neither case breaks the original rule, and again we don't need to turn that card over.

The 7 card is the "if not 3, then not D" case. If we turn over the card and find a D, we've broken the original rule. But if we turn over the card and find some other letter, the original rule has not been broken. We have to turn over that card.

To summarize, of the four cards with D, K, 3, and 7 on the one face, we only have to turn over the D and 7 cards to see if the original rule has been broken for that set of cards. We do not have to turn over the K or 3 cards, their other face does not matter to proving the rule.

Using simple set theory might be easier for some to see these cases.

[Animal]

Thinking about this some more, I think you would have to turn over (3 cards) every single card except the 3.

1. You have to turn over the "D" to make sure they followed the rule and put a 3 on the back.
2. You have to turn over the "K" to make sure that there is not a "D" on the other side, which would make it printed wrong.
3. You have to turn over the "7" to make sure that there is not a "D" on the other side.

But, it doesn't matter what is printed on the other side of the "3". If its a "D" then its printed right. If its any other letter, then it doesn't matter because there is no rule about that.

Close, but still not sparking that bud...

The rule says if a card has a D on one side, then it has a three on the other. So by logic, "if D, then 3." The converse statement, "if 3, then D" is not always true, if the positive is true. "If it is raining outside, you need an umbrella" may be true, but its converse "if you need an umbrella, it's raining outside" is not, as you might need an umbrella for other reasons. The inverse of "if D, then 3" is "if not D, then not 3" is also not always true. "If it is not training, you don't need an umbrella" is not true. On the contrapositive "if not 3, then not D" is true if "if D, then 3" is true. "If you don't need an umbrella, then it is not raining" is a true statement if the original statement is true.

So for "if D, then 3", the converse is "if 3, then D", the inverse is "if not D, then not 3", and the contrapositive is "if not 3, then not D".
That is by the statement of the original rule "if a card has a D on one face, it has a three on the other."
The converse says "if a card has a three on one face, it has a D on the other." -- not necessarily true.
The inverse says "if a card does not have a D on one face, then it does not have a 3 on the other." -- also not necessarily true.
The contrapositive says "if a card does not have a 3 on one face, it does not have a D on the other." -- this is necessarily true.

The cards are chosen to illustrate each kind of statement:

The D face is "if D, then 3" case. If we turn over the D and the other side is 3, the rule is satisfied. If the other side is not a three, the rule is broken and we have a bad set of cards.

The K face is the "if not D, then not 3" case. We turn over the K card and can find either a 3 or some other number, neither case breaks the original rule. So we don't need to turn over the K card.

The 3 face is the "if 3, then D" case. As above, we turn over the 3 card and can find either a D or some other letter, and again neither case breaks the original rule, and again we don't need to turn that card over.

The 7 card is the "if not 3, then not D" case. If we turn over the card and find a D, we've broken the original rule. But if we turn over the card and find some other letter, the original rule has not been broken. We have to turn over that card.

To summarize, of the four cards with D, K, 3, and 7 on the one face, we only have to turn over the D and 7 cards to see if the original rule has been broken for that set of cards. We do not have to turn over the K or 3 cards, their other face does not matter to proving the rule.

Using simple set theory might be easier for some to see these cases.

I am going to have to argue this point. Let's just simply talk about one card. The "K" card. In your explanation, you say that you don't have to turn it over to figure out if the rule has been broken. But let's say there is a "D" on the other side of the card. Has the rule been broken? They printed a card with "D" on one side and did not put a "3" on the other. I say the rule was broken in that case.

UNLESS you are suggesting that there is another rule involved, which states that one side of the card must have a letter and the other side must have a number and its impossible for that rule to be violated.

[QillerDaemon]

I am going to have to argue this point. Let's just simply talk about one card. The "K" card. In your explanation, you say that you don't have to turn it over to figure out if the rule has been broken. But let's say there is a "D" on the other side of the card.

"A card manufacturer makes a set of cards that have a letter on one side of a card and a number on the other side."

That's a given from the original problem. All cards have one letter on one side and one number on the other side.

No card is going to have two letters or two numbers on both sides of each card.

[Animal]

I am going to have to argue this point. Let's just simply talk about one card. The "K" card. In your explanation, you say that you don't have to turn it over to figure out if the rule has been broken. But let's say there is a "D" on the other side of the card.

"A card manufacturer makes a set of cards that have a letter on one side of a card and a number on the other side."

That's a given from the original problem. All cards have one letter on one side and one number on the other side.

No card is going to have two letters or two numbers on both sides of each card.

ah. okay, i missed that part. If that's true, then you only need to look at "D" and "7". K won't matter and 3 won't matter.