Posts tagged with: challenge

It’s been a long time since I’ve done one of these, but I’ve been itching to give away a Math Guide for a while, and I woke up with this question in my head, so I figured I’d let ‘er rip.

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Full contest rules are here, but the important rules are below:

• You must live in the US to get the book for free. If you live outside the US, you must pay shipping. If you are not willing to do that, please don’t enter.
• You cannot have won a Math Guide in a contest in the past.

Assuming there’s a winner tonight (11/12/14), I will update this post tomorrow. Good luck!

UPDATE 11/13/14: Congratulations to Ashish, who made me worry that this question wasn’t hard enough when he got it so quickly. Then a bunch more people got it wrong (about 15% of people got it right during the contest) and I was relieved.

Now that there’s a winner, you can answer the question again (if you already attempted it). This time, when you submit, there’ll be a solution. So if you want to see how it’s done, just try the question again!

How’s everyone else doing on this challenge?

It’s exciting times around PWN HQ—lots of things going on. 2014 should be a fun year for SAT prep. That has nothing to do with this contest, of course. I just like to open these contest posts with a little friendly chatter. I bet nobody even reads this stuff. :/ ANYWAY, here’s a challenge question!

For all positive integers n, let ↭n equal the number of unique prime factors of n. For example, ↭12 = 2, because there are 2 unique prime factors of 12: 2 and 3. If a is a positive integer less than 500,000, what is the greatest possible value of ↭a?

Post your answer in the comments—the first correct answer from someone who hasn’t won a contest before wins a Math Guide! Comments are set to require moderation until tomorrow to add a little suspense into the contest.

All the usual contest rules apply. One recent addition to the contest rules I’d like to draw your attention to:

• No answer changing. When you post a comment, I get an email with that comment, and that’s what I use to judge the contest. Edits applied to your comment later don’t count, even if the edit occurs before someone else wins. Don’t post your comment until you’re sure of your answer.
Good luck!

UPDATE: Wow—I thought this would be tougher for you guys! Congrats to everyone who got it right, especially to Nick, who got his answer in first. Explanation follows below the cut…

The key to getting this question right is recognizing that the smaller the prime factors you use, the more you’ll be able to fit in before your product exceeds 500,000. Therefore, all you need to do to solve this problem is test out the incremental products of the smallest prime numbers!

2 × 3 = 6
2 × 3 × 5 = 30
2 × 3 × 5 × 7 = 210
2 × 3 × 5 × 7 × 11 = 2,310
2 × 3 × 5 × 7 × 11 × 13 = 30,030
2 × 3 × 5 × 7 × 11 × 13 × 17 = 510,510 ← Oops, too big!

What we’ve just shown there is the smallest number that can have 2 unique prime factors is 6, the smallest number that can have 3 unique prime factors is 30, …, the smallest number that can have 7 unique prime factors is 510,510. Therefore, no positive integer less than 500,000 can have more than 6 unique prime factors.

I was just noodling around with Geometer’s Sketchpad today, you know—like a totally normal person, and I made this figure, which I thought would make an awesome challenge question. Wasn’t even planning to post a challenge question, and then BAM! Inspiration.

In the figure above, equilateral triangle DEF is partially obscured by a semicircle with center B and a radius of 5. What is the area of the shaded region?

First correct answer in the comments gets a Math Guide! (Usual contest rules apply.) I’m trying something new this time, though. If I did the settings right, comments will not appear until the contest is over. That way, you can’t see everyone else’s comments before posting yours.

Also, don’t freak out if you’re stumped. My challenge questions are for fun. Although they don’t test anything you don’t need to know for the SAT, they’re usually harder than what you’d find on the SAT.

UPDATE: Congrats to al599, who got it first. Solution below the cut.

Ok. So. This is a shaded region question, and as is usually the case with shaded region questions, we’re going to approach it by finding a larger area, and then subtracting easy-to-calculate unshaded parts. Let’s start by finding the area of the whole equilateral triangle DEF. We know the radius of the semicircle is 5, which means the sides of DEF all have a length of 10.

A good thing for you to remember on the SAT is the fact that all the angles in an equilateral triangle measure 60º. This is important here because it means finding the area of an equilateral triangle is as easy as breaking it into two 30º-60º-90º triangles to find the height.

Cool. So the area of DEF is (1/2)(10)(5√3) = 25√3.

Now we need to start chopping up the unshaded parts of DEF so that we can subtract them and be left with only the shaded part.

Note that if you draw segments from B to the points where the semicircle and triangle meet, you create little equilateral triangles in the lower left and right corners. Those equilateral triangles have 1/4 of the area of DEF, so taking two of them away will take away half the area of DEF. (Take a minute to understand why those are equilateral triangles, and why they’re 1/4 of the triangle. If you’re stuck, ask in the comments.)

So now you’re left with half of the area of DEF, and you still need to subtract the pizza-wedge shape in the center of the semicircle.

To calculate that area, just remember that, because a straight line is 180º, and both gray triangles are equilateral with angles of 60º, the central angle corresponding to that pizza wedge (AKA sector) must also be 60º. We want to find the area of a 60º sector with a radius of 5. A whole circle with radius 5 has an area of 25π, so a 60º sector will have an area of 1/6 of 25π. (For more on how to do this, read this page on circle questions).  Yesterday I went to a museum and waited in line for a very long time for my turn to spend 5 minutes staring at a piece of art that completely mystified me. I guess it’s good art if I’m still thinking about it a day later, even if the thoughts I’m having mostly revolve around my own resignation that I will never really understand art. But it wasn’t a complete loss! There was a circular pattern on the floor that I spent a lot of time staring at while in line, and it ended up inspiring this challenge question.

As always, first correct answer in the comments will win a Math Guide. All the usual contest rules apply: previous winners can’t win; if you live outside the US you have to pay for shipping; etc.

In the figure above, two congruent circles are tangent at point D. Points D, E, and F are the midpoints of AB, AC, and BC, respectively. If AB = 12, what is the area of the shaded region?

Good luck!

UPDATE: Congratulations to John, who got it first. Solution below the cut…

When we’re asked to solve for the areas of weirdly shaped shaded regions, we’re almost always going to find the area of a larger thing that we know how to calculate, and then subtract small things we know how to calculate until we’re left with the weird shaded bit:

The first thing we should do is mark this bad boy up. We know AB = 12, and D is the midpoint of AB and also the endpoint of two radii. We also know E and F are endpoints of two radii, and midpoints of AC and BC, respectively.

At this point, we actually know a great deal. First, we know the radius of each circle is 6. That means each circle has an area of π(6)2 = 36π. We’ll come back to this in a minute.

It should also be obvious that ABC is an equilateral triangle. This is awesome, because equilateral triangles are easily broken into 30º-60º-90º triangles, which is what we’ll do to find the triangle’s area.

So triangle ABC has a base of 12 and a height of 6√3.

Now that we have that, all we need to do is subtract the areas of the circle sectors (in green below) that aren’t included in the shaded region.

Areas of sectors are easy to calculate. All we do is figure out what fraction of the whole circle the sector covers by using the central angle. In this case, the angles are 60º, so we’re dealing with 60/360 = 1/6 of each circle.

We need to subtract two sectors from the area of triangle ABC to find our shaded region:

And there you have it! Cool, right?

This morning I was lying in bed waiting for my alarm to go off, and a question just popped into my head. This happens to me fairly often, but usually I’m too lazy to write it down, then I forget it, then I spend the rest of the day trying to convince myself that it wasn’t a very good question anyway so I shouldn’t worry about forgetting it. This morning, though, I wrote it down. Which means someone’s going to win a Math Guide.

As always, the first correct response in the comments will win the book. If you submit your comment and it doesn’t show up right away, don’t freak out and submit a million more. There’s sometimes a delay because of comment moderation, but rest assured that I receive all comments in the order they’re submitted. If you’re the first one who got it right, I’ll know. (Full contest rules.)

Let’s do this.

An equilateral triangle is inscribed in a circle with area 9π. What is the area of the triangle?

Good luck!

UPDATE: Rob got it first! Solution below the cut.

OK…start by drawing it:

If the area of the circle is 9π, then its radius must be 3 (A = 9π = πr2; 3 = r):

Since you’re dealing with an equilateral triangle, you know each angle in the triangle measures 60º. Since we just made a bunch of smaller, isosceles triangles in there, each of the smaller angles must measure 30º. Which means the central angles must each be 120º.

We want the area of the equilateral triangle, so we need to find its dimensions. To do this, we’re going to use the properties of 30º-60º-90º triangles.

The 30º-60º-90º triangle on the bottom right has a hypotenuse of 3, which means its short leg must be half of that (1.5), and its long leg must be 1.5√3.

Of course, that means the height of the equilateral triangle is 4.5, and its base is 3√3.

Now that we have the base and height, we can calculate the area of the triangle.

Note that although this is a challenge question, probably a tiny bit harder than you’d find on an SAT, it’s quite useful to be able to break an equilateral triangle up into 30º-60º-90ºs to find its area. This is something worth practicing. Phew! It’s only been like three weeks since I last posted but in Internet time that’s forever. Sorry about that. I’ve got about two weeks left in my Master’s program, and I’ve been completely inundated by papers and presentations. My thesis is basically eating me alive. I seriously don’t know how I’m going to get it all done. But that’s school, right? Yay, school!

But I’m not posting to complain! I’m posting because an awesome challenge question popped into my head last night, and because I wanted to tell you what I’m planning to do as soon as school is over and I have more than a few minutes a day to devote to SAT pwning.

Coming soon
• A print version of the Essay Guide
• Some other cool things I’ve been dreaming up that I don’t even know what to call yet
• Some videos maybe? I dunno.

The point is that things are going to pick back up around here. Soon.

And now that challenge question

As always, first correct response in the comments wins a Math Guide. Full contest rules here, but the important ones are:

• You can’t be anonymous because I need to be able to contact you
• You will have to pay for shipping if you live outside the US
• You can’t win more than once, so please refrain from answering if you’ve won a contest in the past.
• If your comment doesn’t appear on the site right away, don’t panic—you’re just getting stuck in my spam filter because you’re not registered. I receive comments in my email in the order they’re posted.

The sum of n consecutive integers is 1111. What is the greatest possible value of n?

Good luck!

UPDATE: Tutor Delphia (a fellow tutor?!) got it first. Solution below the cut.

This question is, I admit, a bit sadistic. That’s because until the correct solution becomes apparent, it can really feel like you’re being asked to find a needle in trial-and-error haystack. Can you find 2 consecutive integers that add up to 1111? Sure, that’s not so hard. How about 3? How about 4? How about 5? etc.

There are a bunch of consecutive integer sets that will add up to 1111, and there’s no easy algorithm to find them (although I’m sure you could come up with an algorithm if you really put your mind to it). If you’re well-versed in SAT math—and even with my challenge questions I try to adhere to SAT principles—then your spider sense should be tingling. There must be an easier way!

Well, here it is: when you’re dealing with sums of consecutive integers, and that set of consecutive integers includes both negative and positive numbers, then corresponding negatives and positives cancel out. This is obvious, once you think about it a bit:

(–2) + (–1) + 0 + 1 + 2 + 3 = 3

See how the –2 and 2 cancel out, and the –1 and 1 cancel out, leaving only the 0 and the 3? The greatest number of consecutive integers that sum to 3, it turns out, is 6. Cool, right?
But what happens if we add another consecutive integer?

(–2) + (–1) + 0 + 1 + 2 + 3 + 4 = 7

Now we’ve got a set of 7 consecutive integers with a sum of 7. Is that the greatest number of consecutive integers with a sum of 7? No. Not at all:

(–6) + (–5) + (–4) + (–3) + (–2) + (–1) + 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 = 7

That’s right—we can construct a set of 14 consecutive integers with a sum of 7. In general, the largest number of consecutive integers that sum to a given positive integer will be twice that integer.

Therefore, the greatest number of positive integers that sum to 1111 is 2222:

(–1110) + (–1109) + … + 0 + … + 1109 + 1110 + 1111 = 1111

There are 1110 negative integers, 1111 positive integers, and 0.

1110 + 1111 + 1 = 2222

You know, It’s funny. When I sit down to write one of these, I usually have a rough sketch in my mind of how the question’s going to go. But half the time, the problem completely evolves into something else by the time I’m done with it. I’m not sure why that is. It could just be that I’m capricious, but it’s probably because math comes into much sharper focus once it’s written down. Which I only mention to gently encourage you to write everything down when you take tests.

Anyway, you know the drill by now. First correct response in the comments wins a copy of my Math Guide. Full contest rules here.

Note: If you’re new to the site, your comment might not appear right away. Don’t panic—I receive comments in my email in the order they’re submitted. If you’re the first to get it right, you’ll win even if your comment doesn’t immediately appear.

In the figure above, A is the vertex of f(x), B and C are the x-intercepts of f(x), F is the origin, D is the midpoint of AF, and E is the midpoint of DF. If f(x) = px2 + 8 and the area of concave quadrilateral BDCE is equal to 8, what is p?

Good luck!

UPDATE: Peter got it first. Solution below the cut.
The first thing to focus on here is the equation, f(x) = px2 + 8, which tells you a very important thing: the y-intercept of f(x) is 8. This is one of the few things you need to keep in mind about parabolas—that the constant in a parabola equation tells you its y-intercept. This makes sense, of course, because the y-intercept is when x = 0, and f(0) = p(0)2 + 8 = 8. So yeah. Point A is (0, 8).

We know that D and E are midpoints, so we now know a few distances. Let’s throw them on the figure:

The only other thing we have to work with is that BDCE has an area of 8. There are a few ways one might use that information—and please, as always, leave your own solutions in the comments—but I recommend recognizing that, because parabolas are symmetrical, everything on this figure is symmetrical about the y-axis. So if BDCE has an area of 8, then △DEB and △DEC each have areas of 4. We can work more easily with triangles.
Let’s focus on △DEC.

If △DEC has an area of 8, and a base of DE = 2, we can solve its height, FC.* We do this with the handy-dandy triangle area formula:
Anyway, now we know FC = 4. Since F is the origin and C is on the x-axis, that means point C is (4, 0).
And we can use that to solve for p. All we need to do is plug the point we know is on the parabola, C, into the parabola’s equation. We know (4, 0) is on the parabola. (Another way of saying this: f(4) = 0.)
f(x) = px2 + 8
f(4) = p(4)2 + 8 = 0
16p + 8 = 0
16p = –8

p = –0.5

And there you have it! Pretty awesome question, if I do say so myself. * If you’re unhappy about me calling DE the base and FC the height, have a look at the same triangle rotated 90º. Better? Cool.

It’s been a while since I’ve posted a challenge question. There are two main reasons: my own schoolwork, and the fact that most of my PWN time in the past month or so was dedicated to the Essay Guide. Now that I’m done with that, it’s time to get back in the swing of things with some pull-your-hair-out, way-harder-than-the-SAT-but-work-the-same-muscles math questions!

As always, the first non-anonymous commenter to get this right will win a copy of my Math Guide. Note that I’ll only count your first answer, which is emailed to me when you submit your comment—editing later doesn’t work. Don’t submit until you’re sure! (Full contest rules.)

Let’s do this.

The figure above shows the graph of f(x). If a is an integer such that
–6 < a < 6 and f(a) = a, what is f(f(f(a – 3)) + f(f(f(f(2)))))?

MUAHAHAHAHA. Good luck!

UPDATE: It’s over! This one fooled a bunch of people. Solution below the cut.

There are a few steps here. I don’t think many of you had trouble with the first one, which is to figure out a. Here’s one an important thing to remember about function notation: you can translate function notation expressions like f(p) = q into ordered pairs (p, q). So when the question says f(a) = a, you’re looking for the ordered pair (a, a). There’s only one point on the graph where that occurs: (–2, –2). So a = –2, and the expression becomes:

f(f(f(–2 – 3)) + f(f(f(f(2)))))

or…

f(f(f(–5)) + f(f(f(f(2)))))

From there, this question is really just an exercise in reading the expression carefully, and finding points on the graph. I think where a lot of people got turned around was that they didn’t work from the inside out, and therefore misread how the parentheses grouped the expression. Let’s go through it together, one step at a time.

f(f(f(–5)) + f(f(f(f(2)))))

Look to the graph to find f(–5) and f(2). To do this, follow the x-axis over to –5 and 2, and find the y-coordinate of the graph at those points:
So f(–5) = 2 and f(2) = 3. Now just repeat that same process a few more times, working from the inside out.
f(f(2) + f(f(f(3))))

Uh oh. Look at that. Once we take care of the f(2) on the left there, we run out of closed parentheses on the left. This does not mean I made a typo writing the question. Rather, it means we have to do more work on the right to close off the first parentheses. Like so:

f(3 + f(f(0)))
f(3 + f(–3))
f(3 + 1)
f(4)
1

And there you have it—the answer is 1. Fun, right?

My holiday book giveaway is still going strong until January 1, and you can still get your name in there to try to win the daily random drawing, but I figured it was high time for another challenge question. The first person to post a comment on this post with the correct answer will win a copy of the Math Guide. The usual contest rules apply.

Charles has painted a diameter on a wheel with radius of x centimeters, and is holding the wheel in place on an inclined plane so that the painted line is parallel to the ground as shown in the figure above. If he releases the wheel and it begins to roll without slipping, how many centimeters, in terms of x, will it roll before the painted diameter is perpendicular to the inclined plane for the first time?

As I’ve mentioned on the last few challenge questions, some of my older posts have been getting spammed so I’ve had to add some security to the comment system. If your comment doesn’t appear right away, don’t panic. That’s normal. I still receive an email for every comment as it comes in, so I will still know who got the answer first. You don’t need to resubmit 50 times. 🙂

Good luck!

UPDATE: Many congratulations to Muna for getting it first. Solution below the cut.

The key to this question is really just figuring out how many degrees of rotation the wheel will make for the painted diameter to be parallel to the ramp. To figure that out, note that right now, it’s at a 30º angle from the ramp:

It’s going to roll downhill, obviously, so the first 30º it rolls will make the painted line parallel to the ramp:

From here, it just needs to roll 90º further to make the line perpendicular to the ramp. Total revolution: 30º + 90º = 120º.

Of course, 120º is 1/3 of 360º, so the wheel is going to make only 1/3 of a revolution.

Circles, as you probably know, roll a distance of one circumference when they make one complete revolution. Since this wheel only makes 1/3 of a revolution, it’s only going to roll a distance of one circumference.

Someone requested recently at my Q&A page that I post a challenge question about direct and inverse variation. I quickly agreed, but it’s taken me longer than I wanted it to because I’ve been having a hard time hitting the right level of difficulty for a challenge question. I’m still not sure I’ve nailed it, but regardless, the wait is over. First correct response, as usual, gets a Math Guide shipped to their home (in the US) for free. Full contest rules, as always, apply.

• p and q are directly proportional to each other; their proportionality constant is k
• x and y are inversely proportional to each other; their proportionality constant is j
• k = j
• a is a constant greater than 1

According to the conditions above, what is one possible value of y in terms of a when p = a3, q = a, and x = a?

Put your answers in the comments. If you’re not registered with this site and your comment doesn’t appear immediately, don’t panic. I get comments in the order they’re submitted, but not everything shows up right away because I’m trying to prevent spam.

UPDATE: Rushil won the book, and Peter gave a nice explanation in the comments. My solution is below the cut.

To begin work on this, get all the information in the bullet points into equations.

If p and q are directly proportional, with proportionality constant k, there are actually two equations you could write:

This is because all that really needs to be true is that p and q always make the same fraction. I’m going to deal with the first one first, and then I’ll give a quick treatment to the second one. I would have accepted the answer that results from either one.

If x and y are inversely proportional, with proportionality constant j, there’s only one equation you can write:

Of course, if k and j are equal, then we can set up one equation:

And now we’re ready to substitute in our values. Remember, p = a3, q = a, and x = a. So here we go:

And now we solve for y:

Had you used the second version of the p and q equation, you’d instead get y = a–3 here instead. I would have accepted that as an answer as well. Here’s how that’d work:

Someone on the Q&A page asked me to do one of these “point A to point B” questions, so I figured I’d try to come up with one worthy of being called a challenge question. I think this is a bit easier than the usual. I expect someone to get it pretty quickly.

Please note: This is the first Challenge I’ve posted since I made some slight changes to the commenting policy on this blog. If you don’t have a Disqus account with a verified email address, your comment will not appear on the site immediately. Don’t freak out—I get an email for every comment that’s posted. I will still be able to tell who responded correctly first, so I’ll know who to award the Math Guide to. As always, the usual contest rules apply.

Joss is playing an elaborate version of the game everyone plays as a kid where the floor is lava and you can only step on pillows*. In Joss’s version of the game:

• The pillows are laid out in a pattern just like the one in the figure above
• Joss can only move the way a knight in chess moves (so his first move will put him 1 diagonal space away from a red corner)
• Any pillow he lands on disappears when he leaves it (assume he leaps directly to the pillow his move would end on, and does not touch other pillows in between)

If Joss begins on the center pillow, and makes the minimum number of moves necessary to land on each red pillow before leaping to safety on the couch, how many pillows will remain?

*I know not everyone has played this game. Hopefully, even if you haven’t, you can picture the scene.

UPDATE: Props to Peter, who got it first. Solution below the cut.

I’ve never posted a video here before (which is saying something considering that between this main blog and the Q&A Tumblr I’m well over 1500 posts) but this question really seemed to lend itself better to a video explanation than a written one.

I didn’t want to re-record the whole thing to fix this, but note that even though I described a knight’s movement as “up two, over one,” I am aware that a knight can also move down. Anyway, here’s the solution:

Thanks to Hurricane Sandy, this has been a rough week for a lot of people. If you’re feeling charitable, please note that I am still giving away Math Guides for donations to the Red Cross.

Let me note, before we get into this, that these challenge questions are WAY harder than anything you’d see on the SAT. That’s why I call them challenges. They test your knowledge of concepts that appear on the SAT, sure, but my challenge questions are meant to stump you for at least a few minutes.  Please don’t freak out about these, especially if you’re taking the November SAT tomorrow. Anyway, let’s get to it.

Amy is putting together an epic 7-song playlist that she can listen to while she’s working out at the gym. She’s a bit particular about how it’s constructed. The second song must be 25% longer than the first. The third song must be 12.5% shorter than the second. The fourth song must be half a minute shorter than the third. The fifth song must be 1.5 times as long as the fourth, and the sixth song must be 30% shorter than the fifth. The seventh song must be her current favorite song, “Gangnam Style,” the mp3 for which she recently purchased from Amazon. The total playlist must last exactly 25 minutes, with no pauses between songs. How long must the first song on Amy’s playlist be?

As usual, the first correct answer in the comments will win a copy of the Math Guide.

[Full contest rules.]

UPDATE: It’s taken me a long time to write the solution to this, because, well, I kinda started thinking I had already done it. Commenter Yeana got it right first, and the commenter Peter posted a good short explanation. My full explanation below the cut.

The first thing I’d recommend doing is deciding whether you want to work with minutes in decimal form, or in seconds. My preference is for seconds (and I went to some trouble to make sure none of these songs had fractional second lengths). So yeah. My explanation will use seconds.

A 25 minute playlist will be 25 minutes × 60 seconds per minute = 1500 seconds long. And “Gangnam Style” (I specified the Amazon version because I found other online versions that differed by a second and I didn’t that to frustrate anyone) is 3 minutes and 39 seconds, or 219 seconds long. Everything else you’re going to have to figure out by reading through the problem.

Let’s work through this step by step.

• Since the question asks about the first song, let’s call its length x.
• The second song’s length is 25% greater than that of the first. So it’s going to be 1.25x.
• The third song is 12.5% shorter than the second. Another way to think of this is that it’s 87.5% as long as the second. So the third song will be 0.875(1.25x) seconds long.
• The fourth song is an easy one, but at the same time continues to complicate the question. It’s 30 seconds shorter than the third song, so its length is 0.875(1.25x) – 30.
• The fifth song is 1.5 times as long as the fourth, so it’s 1.5(0.875(1.25x) – 30) seconds long.
• The sixth song must be 30% shorter than the fifth (in other words, 70% as long as the fifth). So its length is 0.7(1.5(0.875(1.25x) – 30))
• And of course, as already stated, the seventh song is 219 seconds long.
All of those add up to 1500. So let’s solve for x!
…On second thought, let’s let Wolfram Alpha do it for us. 🙂
x = 192
So the last thing we need to do is convert that to minutes and seconds. Note that seconds-to-minutes conversions are one place other than the SAT and 3rd grade where remainders are useful. The closest minute mark is 3 minutes (180 seconds), and there are 12 seconds left over. In other words, 192/60 = 3 remainder 12. So the first song is 3:12 long.

Since you guys tore up my last challenge question so quickly, I figured I’d make this one a tiny bit harder. Let’s see if I can’t stump you for more than 8 minutes.

First correct response in the comments gets a free Math Guide. Usual contest rules apply. Ready? Here we go!

In the figure above, triangle ABC is inscribed in a circle with center P. If the area of triangle ABC is equal to , and AC = xy, what is the area of the circle in terms of x and y? (Be careful typing your answer—parentheses matter!)

MUAHAHAHA. Good luck!

UPDATE: Tuấn got it first, although a others came VERY close (careful with parentheses!!!). Solution below the cut.

It might not be immediately clear what to do here, but as is true of most geometry questions, it’s a good idea to start by listing everything you know. First, you know that triangle ABC is a right triangle, because BC is a diameter, and point A is on the circle. You might recall from geometry class that if an angle lies on a circle, its measure is half of the measure of the arc it corresponds to. Since A is on the circle and corresponds to a 180º arc, the measure of angle A is 90º. You probably won’t see that rule tested on the SAT, but it’s not outside the realm of possibility. So here’s what we know:

Of course, we also know that the formula for the area of a triangle is ½bh. So let’s figure out AB:

Oh look. Difference of two squares. Crazy how often that appears!
Now we can solve for BC, using the Pythagorean Theorem:
Nice. Here’s what we’ve got now:
Of course, BC is the diameter of the circle, and to find the circle’s area we want the radius, which is half of BC:

Hey guys. Hope you’ve been well. I’ve been buried under mountains of work for grad school. It’s awesome. I’ve also been answering a ton of questions over at my Q&A site, which is really awesome, minus the sarcasm inherent in the previous usage of “awesome.”

Anyway, I was thinking about you guys the other day, and how I haven’t given away very many books lately. I want to remedy that. I like giving away books. So, first person to answer the question below correctly in the comments (non-anonymously) gets a Math Guide, free of charge (full contest rules here).

In the equation above, a, b, c, and d are positive integer constants, and
a > b > c > d. What is the least possible value of a?

Yeah. I didn’t feel like going easy on you guys for this first Weekend Challenge in a while. The winner of this book will have earned it. 🙂

UPDATE: Well, that was fast. “DE” got it 8 minutes after the post went up. Guess I’ll have to make the next one harder! Solution below the cut.

The first thing you’re going to want to do here is manipulate the numerator of this fraction:

Hmm…we can factor those:

Now here’s where it gets awesome. That’s a difference of two squares! Do you see it? It’s one thing squared minus another thing squared. Doesn’t matter that each thing is more complicated than a single variable. So we can factor:

Simplify:

Simplify some more:

And there you have it. a – b – cd = 0. Since we have the constraints that all four values are positive integers and that a is the greatest of them, we have enough information to solve for the smallest possible value for a.

To get the smallest possible a, we need to start with the smallest possible d, which would be 1. If d = 1, c = 2, and b = 3, then a is the smallest is can possibly be.

I’m going to be out of town for a few days, so I won’t be able to post a solution for this until probably Monday, but hopefully I’ll be able to check in and at least congratulate the winner, who will receive a free copy of my Math Guide.

A student is drawing concentric circles, each with a diameter one centimeter longer than that of the one before. When he tires of this random task, he has drawn 21 circles. If the area enclosed by his circles increased by 300% from the time he drew his first circle until the time he was done, what is the radius of the smallest circle?

LATE UPDATE: David’s the winner—he got the correct answer first. I’ve been buried under a bunch of other things and have been slow to get the solution posted. Sorry about that. It’s below the cut.
The first task here is to figure out how much bigger the last circle is than the first. This requires a tiny bit of care. Note that the second circle’s diameter is 1 cm bigger than that of the first. The 3rd circle’s diameter is 2 cm bigger than that of the first. And so on. So the 21st circle’s diameter will be 20 cm larger than that of the first.

Of course, this means the radius of the largest circle is 10 bigger than the radius of the smallest circle, which is what we really care about if we’re talking about areas.

Let’s call the radius of the smallest circle r. The area of the smallest circle is πr2. The area of the largest circle is π(+ 10)2.

The next tricky part is the 300% increase. An increase of 300% does not mean that the new area is 3 times bigger than the old. It means the area increased by its original size 3 times over.

This is an important insight, and it makes this problem a lot easier. I’ll include the mathy way below, because I like that kind of thing and I think it’s important to show that there are multiple ways to solve this, but note that it’s a bit more involved if you don’t know that a 300% increase means you end up with 4 times what you started with.

If the new area is 4 times the original area, we can set up a simple equation:

At this point, you’re going to have to use the quadratic formula, or do some factoring. You never need the quadratic formula on the SAT, so I’d recommend factoring, but really it’s whatever makes you happy. 🙂

You know your first terms have to be 3r and r, and you know one factor must contain subtraction to get -100 in the last term, etc. A little trial and error later, you get…

So your solutions are r = 10, or r = -10/3. Since a circle’s radius can’t be negative, the answer is r = 10.

That was a bit mathy, right? Here’s the other, slightly mathier way:

If the area enclosed increased by 300%, then we can use our percent change formula to solve.

Let’s clean that up a little bit by dividing both sides by 100% and knocking the π term out…

And simplify a little more…

And that’s the same place we ended up at above. You’ll still get r = 10.