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:

AwholeAunshaded = Ashaded

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?

Comments (9)

From guesstimating, it looks like the non shaded parts of the circle is
1/6 of the circle. Give that AB=12 with midpoint D, then AD and DB each
have a value of 6.
π(6)^2 = 36π per circle area. 1/6 of that is equal to 6π. 6π+6π= 12π.
Triangle is equilateral with altitude, 6√ 3. So, (12 x 6√ 3)/2 = 36√ 3
36√ 3 = area of triangle 36√ 3 – 12π= roughly 24.655

I also got 36√3 – 12π.

Since D is the midpoint of AB, AB=12, AD = DB = 6 by the definition of a midpoint. Since E and F are the midpoints of AC and BC, we know that each side length of the triangle is 12. Recall that the area of an equilateral triangle with side length s is s^2√3 / 4. (This can be proven by creating drawing 6 congruent 30-60-90, and finding the area in terms of s). From the formula, the area of the triangle is 36√3. Now the two sectors we subtract are each 1/6 * π * (6)^2 = 6π. Since there are two, the area of the two are a combined 12π. Hence, the area of the shaded region is 36√3 – 12π.

Please explain question 18 given on page 132 (PWN the SAT) under parabola questions.

the explanation given in the book it is said that x coordinates of
points MUST be equidistant from x=6, but why?? and it is also said that y
coodinate MUST be same but why??

do we ever need to use equations solving for parabola question on SAT ??

The question tells you that the parabola’s minimum is at x = 6, which means its line of symmetry is x = 6. From there, you’re looking for points on either side of that line that could match the shape of the parabola.

Since you know it has a minimum at x = 6, y-values must increase the farther points get away from x = 6. That eliminates (A), (B), (C), and (E), because each of those has either two points the same distance from x = 6 but with different heights, or two points where the one that’s farther from x = 6 is lower than the one that’s closer, which couldn’t match the shape of a parabola that opens upwards.

Great, thanks!!
But here is another one coming….
question 18 page 210..

.I get the answer by listing the outcomes (as you have explained on solutions page)

But, I want to know the mathematical way of getting the answer.

Yes, listing make the answer attempt a sure shot BUT think if this type of question is the last question in the section and proctor is about to ask to put the pencils down, it would be better to attempt using mathematical method.

On your website, I saw for the first time that author of the SAT book is so engaged with students preparing for SAT.

I am glad you are finding the book and the site useful. 🙂

I don’t think the “math” solution to that question is particularly intuitive, but here goes:

First, establish the number of possibilities based on the number of office: 4 × 3 × 2 × 1 = 24

Then, note that if those two need to be next to each other, you can pretty quickly count how often that will be possible. Say the two guys we care about are XX, and the other employees are o’s:


Those are the only 3 ways the friends can be next to each other. Of course, for each of those we need to multiply by 2 because Scooter 1st and Big Man 2nd is a different arrangement than Big Man 1st and Scooter second. Then we need to multiply each choice by 2 AGAIN because the same is true of the employees we don’t care about—they could be in two orders. In other words, if the two guys we didn’t care about were Ed and Frank, XXoo could be SBef, BSef, SBfe, or BSfe.

So there are 3 × 2 × 2 = 12 ways the friends could be next to each other. 12/24 = 1/2.

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