The Modern GMAT’s Ancient Roots
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Mathematics credits its most famous formula to a legendary Greek, Pythagoras of Samos. Although Babylonians and Indians had used the formula centuries before his birth, this Greek scholar is widely known as being the first to prove the theorem.
Recently, however, historians have contested whether he was truly the first to discover the formula. Part of the debate arises because Pythagoras and his pupils were part of a secretive school of mathematics which left no written records — and hence no evidence.
Yet regardless of who ultimately discovered the theorem, this gem of trigonometry has had far-reaching impacts in every sphere of mathematics. Surprisingly, it remains simple enough to teach to children in middle-school:
Pythagorean’s theorem applies to any right triangle (a triangle that contains a 90-degree angle). In the equation, a and b represent the two legs (the shorter sides), and c represents the hypotenuse (the longer side, opposite the right angle).
Memorize these common Pythagorean triples
Although every right-triangle must satisfy this equation, a few have the special distinction of having all their sides contain only integers. These special 3-integer combinations are known as Pythagorean triples, and they appear quite frequently on the GMAT. The most common example is the 3-4-5 right triangle.
A new Pythagorean triples can be formed simply by multiplying an existing triple by a constant number. For example, if we multiply the 3-4-5 ratio by 2, we get the new Pythagorean triple, 6-8-10. Multiples of existing ratios can easily be calculated, so the only triples you need to memorize for the exam are these three unique ratios: 3-4-5, 5-12-13, 8-15-17.
Having learned the necessary background, let’s see some applications of this theorem. Consider the following data sufficiency problem:
In the figure below, what is the value of c/a?
(1) a is 25% shorter than b.
(2) b is 20% shorter than c.
To have sufficient data, we must have one and only one possible value for the ratio of c/a. We are given a right triangle, which means that the variables must obey the Pythagorean theorem. Perhaps if we can use the statements to replace the variable b, we might be able to express the equation in terms of only c and a alone. That will probably get us closer to finding the ratio.
If we translate statement 1), we have a = b – 0.25b = 3b/4. Rearranging, we have b=4a/3 . If we substitute this into Pythagorean’s theorem, we have:
Typically, when dealing with the squares of variables, we should consider both negative and positive solutions. In a geometry problem, however, lengths must always be positive numbers, so we know that both a and c can never take on negative values. c/a is therefore simply 5/3 — it is a single value, so we therefore have sufficient data.
We can apply the same technique to statement 2). Translating the statement and rearranging, we get b=4c/5. We substitute this back into Pythagorean’s theorem to find that:
Here, too, c/a = 5/3. Again, we have a single value, so we have sufficient data. Each statement alone provides sufficient data.
Our quick review of the Pythagorean theorem should have helped you get caught up on GMAT trigonometry. Before we finish, try to use what you’ve learned to solve the following challenge problem:
What is the ratio, by length, of AC to BC?
Right triangle ABC is divided into five identical right triangles as shown above. What is the ratio, by length, of AC to BC?
11 comments
sriram on May 31st, 2012 at 3:07 am
is the answer to the question (sqroot5)/2 ?
thanks you for the good question
PP on May 31st, 2012 at 5:33 am
sqroot of 5/2
Ankita on May 31st, 2012 at 7:48 am
is the Answer =sqr root(5)/2
harsh Singla on May 31st, 2012 at 6:41 pm
please explain ..not able to get it
sriram on June 1st, 2012 at 1:32 am
lines with identical colour are equal in length since right triangles are identical.
this helps us derive that BE=2AE.
this info is sufficient to solve the problem i guess.
ALESSANDRO MASTRACCO on June 5th, 2012 at 10:30 am
Let say:
AB= a
BC=2AB= 2a
AC= b+b+c= 2b+c
AC/BC= (2b+c)/2a
We know from Pythagorean’s theorem that:
(2b+c)=sqr root(aXa+2aX2a)
Then :
(2b+c)/2a=(sqr root(5)Xa)/2a= (sqroot5)/2
ALESSANDRO MASTRACCO on June 5th, 2012 at 10:29 am
Let say:
AB= a
BC=2AB= 2a
AC= b+b+c= 2b+c
AC/BC= (2b+c)/2a
We know from Pythagorean’s theorem that:
(2b+c)=sqr root(aXa+2aX2a)
Then :
(2b+c)/2a=(sqr root(5)Xa)/2a= (sqroot5)/2
Too easy!!!
lucky on July 8th, 2012 at 3:03 am
hey 45,45,90 is also a right angle triangle with two sides equal so rather than considering a,b,c as sides if i'll take sides as (AB=squareroot(2)a,BC=2AB,AC=3a) as there in case of 45,45,90 my answer will vary why so? it must be same in each case please explain?
Kunal on June 7th, 2012 at 8:23 pm
Hi Alessandro: Ur method is too cool..! Thanks
Kristen on September 17th, 2012 at 1:31 pm
Here's a non-algebraic way to get the same answer...
The important relationship is that the shortest side (Green) is equal to exactly half of the second-longest side (Red). We can see this relationship in line BE. (This is why the triangles can't be 45-45-90s, as someone commented.)
Since we just need the ratio, we can use numbers. If we say Green = 1, then Red = 2. From the Pythagorean theorem, Blue = sqrt(1^2 + 2^2) = sqrt(5)
So:
AC = Green + 2(Red) = 1 + 2(2) = 5
BC = 2(Blue) = 2(sqrt(5))
AC/BC = 5/(2(sqrt(5))
This simplifies to (sqrt(5))/2:
5/(2(sqrt(5)) = (1/2)*(5/(sqrt(5))
= (1/2)(sqrt(5))
= (sqrt(5))/2
Vinayak Lahoti on December 29th, 2012 at 5:42 am
I am in grade 10 and i solved this sum in 3 different ways .The answer i got is square root of 5 divided by 2.Excellent problem