Euclid is credited as the founder of most of our modern geometry. Though he lived approximately 300 years B.C., his foundation of geometry is still the basis of the geometry we teach in high school and, therefore, the basis for SAT questions on geometry. One of the most common geometry questions on recent SAT’s goes right back to one of Euclid’s most basic ideas.

What is the shortest distance between two points? Euclid said the shortest distance between two points is a straight line. This was one of his fundamental axioms, or postulates – statements that are accepted without proof because they are so obviously true.

Consider this situation: You need to travel from point A to point B.  According to Euclid, the shortest distance would be along the straight line AB. However, now let’s suppose that you had to divert first to a point C that is not along this straight path. You would first travel from A to C (AC) and then from C to B (CB). Clearly, this second trip which takes us first to point C is the longer trip. If you draw a triangle to represent this situation, we can rename the result: AC plus CB must be longer than directly from A to B, namely AB. Thus, based on Euclid’s axiom, we have proven a theorem: The sum of two sides of a triangle must be greater than the third side.  Someone at the Educational Testing Service has decided that this is a critical concept and a question on it has appeared on most recent SAT’s.

Problem: “Two sides of a triangle are 5 and 9.  If the length of the third side is an integer, find one possible length of the third side.”

Let’s try 3.  This doesn’t work since 5+3=8 which is not greater than 9. Perhaps we should try 4.  Now, 5+4=9 but is not greater than 9 so this is not good. If we try 5, the sum of any two sides is greater than the third side. Note that 6,7,8,9,10,11,12,13 also work. When we reach 14, 5+9=14 but is not greater so that doesn’t work.

Problem: “Two sides of an isosceles triangle are 2 and 5. Find every possible perimeter of the triangle.” Isosceles triangles have two equal sides. So, the triangle could have sides of 5, 5 and 2 giving us a perimeter of 12. However, many students would then suggest that there’s another isosceles triangle with sides of 2, 2 and 5. Sadly, they would be wrong. Since the sum of any two sides must be greater than the third side and 2+2 is not greater than 5, this second triangle doesn’t even exist. The only possible perimeter is 12.

The sum of any two sides of a triangle must be greater than the third side. Make sure you know this when you prep for the SAT.

Every student knows how to find his/her test average. If you’ve taken 5 tests, just add them up and divide by 5 to find the average, or mean. Such a straightforward problem simply won’t appear on an SAT. Instead, one of the best SAT strategies involves how to deal with the “backwards” situation.

Consider this problem: “Three numbers have an average of 50. Two of these numbers have an average of 40. What is the third number?” If three numbers have an average of 50, what must their sum have been? Of course, this sum must have been 3 times 50, or 150. You should train yourself to jot down 150 even before reading the second sentence. This is always the key to such a problem on averages. Use the same method with the second sentence. If two numbers have an average of 40, their sum must have been 2 times 40, or 80. Now, to find the missing third number, just subtract 150 minus 80, yielding 70.

This strategy can be used in a very common type of problem on averages. “On her first 5 math tests, a girl has had test scores of 88, 93, 84, 87 and 89. What must she score on the sixth test in order to pull herself up to a 90 average?” Employing this method, we ask ourselves how many total points must she have after 6 tests in order to have a 90 average. Simply multiply 6 times 90, or 540 points. Adding up her first 5 test scores we see that she had accumulated 441 points. Just subtract 540 minus 441 telling us that she better do some serious studying since she needs 99 points on this sixth test.

Approximately 30% of the SAT math questions concern arithmetic concepts – no algebra or geometry involved. One of the favorite arithmetic topics is averages and this method is often used once or twice on every SAT. I often tell SAT Prep Course students that if the first sentence of the first math problem of their SAT says “Five numbers have an average of 8,” they should immediately write 40 (5 times 8 ) on their paper and this will surely lead to an easy solution.

Colored banners were strung in the following pattern: red, blue, green yellow, white, red, blue, green, yellow, white, and so on. If this pattern continues, what is the color of the 43^rd banner?  

Certainly, if you’re trying to get an SAT problem right by any means, if the number is small enough, just write out the pattern until you reach the 43^rd color. However, what’s the real mathematical solution? First, notice that the color pattern repeats every 5 colors. Divide 43 by 5. Most students would do this on their calculators giving them 8.6. Is that .6 the remainder? NO!!!  We want the remainder and remainders are found at the bottom of long divisions. If you do 43 divided by 5 on paper, you will get a quotient of 8 and a remainder at the bottom of the division of 3. What does this tell us? There were 8 complete cycles, or patterns, of 5 colors, and 3 “left over.” Just count to the third color, green, and you have your answer. 

Another example: We have the repeating decimal .253698253698… and so on. What is the 50^th digit of this repeating decimal? First, we notice that the digits repeat every 6 digits so we divide 50 by 6. The quotient is 8 and the remainder is 2. This tells us that there will be 8 complete cycles of 6 digits and 2 is the remainder. This means that the second digit, a “5”, will be the 50^th digit (not the “2”, this is the first digit of the pattern and it would have been the answer if the remainder had been 1). 

All “cycle” or “pattern’ problems are solved by this same procedure: Identify how often the pattern repeats itself, then divide by this number, and the remainder will always give you the answer.

Consider this grid-in problem:

If 3x + 2y = 16, what is the value of 9x + 6y?

Suppose you don’t see the solution at first. Since it’s a grid-in problem, there are no choices to try so you might think that you are out of business (remember you do get a penalty-free guess on a grid-in question). Is there anything else you can try? Yes, try plugging in some numbers. The equation 3x + 2y = 16 has an infinite number of solutions. For instance, x=2 and y=5 are “good” numbers since 3 (2) + 2 (5) = 16. Now, just take those values and plug them into 9x + 6y giving you 9 (2) + 6 (5) = 48. Grid the answer 48 and you get your one point for this problem. What was the “pure solution to the problem? Ask yourself how the expression 9x + 6y was chosen. It is precisely 3 times the original expression 3x + 2y or 3 (3x + 2y) = 9x + 6y. Just calculate 3 * 16 = 48 and you’re done!