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Geometry Trisected

by mevicks » Fri Sep 20, 2013 4:37 am
Image

In the figure above, ABCD is a square, and the two diagonal lines divide it into three regions of equal area. If AB = 3, what is the length of w, the perpendicular distance between the two diagonal lines?

A) 3√2 - 2√3
B) 3√2 - √6
C) √2
D) Image
E) 2√3 - √6

Time yourself when solving this, and do post your time ! In exam conditions it took me 3.2 Minutes and then I had to guess and move on !
After reviewing the problem in untimed conditions I found out that there are a few "very easy under 2 minutes" solutions available.

[spoiler]OA A : 3√2 - 2√3[/spoiler]
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by GMATGuruNY » Fri Sep 20, 2013 4:57 am
mevicks wrote:Image

In the figure above, ABCD is a square, and the two diagonal lines divide it into three regions of equal area. If AB = 3, what is the length of w, the perpendicular distance between the two diagonal lines?

A) 3√2 - 2√3
B) 3√2 - √6
C) √2
D) Image
E) 2√3 - √6
Image

The area of square ABCD = 3² = 9.

Since the square is divided into 3 equal regions, the area of triangle DWY = 3.
Thus:
(1/2)(DY)(WD) = 3.
Since triangle DWY is isosceles, DY=WD.
Thus:
(1/2)(WD)(WD) = 3
WD² = 6
WD = √6.

Since AD = 3 and WD = √6, AW = 3 - √6.
Triangle AXW is a 45-45-90 triangle.
(Don't worry about a proof. It should be clear from the figure that AXW is an isosceles right triangle.)
Since the sides in triangle AXW are proportioned 1:1:√2, we get:
WX = (3 - √6)√2 = 3√2 - √12 = 3√2 - 2√3.

The correct answer is A.
Last edited by GMATGuruNY on Fri Sep 20, 2013 5:47 am, edited 1 time in total.
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by theCodeToGMAT » Fri Sep 20, 2013 5:23 am
Image

Area of triangle XDC = Area of Sq/3
A^2/2 = 9/3
A = √6
Hypotenuse = √12
Perpendicular from D to XY = (√6 √6)/ √12 => √3
So W = Daignol of sq - Perpendicular = 3√2 - 2√3
Answer [spoiler][A][/spoiler]
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by mevicks » Fri Sep 20, 2013 5:36 am
theCodeToGMAT wrote:Image

Area of triangle XDY = Area of Sq/3
A^2/2 = 9/3
A = √6
Hypotenuse = √12
Perpendicular from D to XY = (√6 √6)/ √12 => √3
So W = Daignol of sq - Perpendicular = 3√2 - 2√3
Answer [spoiler][A][/spoiler]
Nice alternate explanation rahul. Although I have a doubt regarding the step in red. How did you find the value of the perpendicular in one step ? (I assume you have considered the area = 1/2*Base*Height formula to find it).

The explanation by Mitch is the fastest and the easiest though.
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by theCodeToGMAT » Fri Sep 20, 2013 5:46 am
mevicks wrote:
theCodeToGMAT wrote:Image

Area of triangle XDY = Area of Sq/3
A^2/2 = 9/3
A = √6
Hypotenuse = √12
Perpendicular from D to XY = (√6 √6)/ √12 => √3
So W = Daignol of sq - Perpendicular = 3√2 - 2√3
Answer [spoiler][A][/spoiler]
Nice alternate explanation rahul. Although I have a doubt regarding the step in red. How did you find the value of the perpendicular in one step ? (I assume you have considered the area = 1/2*Base*Height formula to find it).

The explanation by Mitch is the fastest and the easiest though.


Mevrick, there's one rule; The rule says:

Image

where H is the Perpendicular on Hypotenuse
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by GMATGuruNY » Fri Sep 20, 2013 6:04 am
mevicks wrote:
theCodeToGMAT wrote:Image

Area of triangle XDY = Area of Sq/3
A^2/2 = 9/3
A = √6
Hypotenuse = √12
Perpendicular from D to XY = (√6 √6)/ √12 => √3
I have a doubt regarding the step in red. How did you find the value of the perpendicular in one step ?
Any side of a triangle can be deemed THE BASE.
Each base has a corresponding HEIGHT.
Since the area of the triangle must be the same no matter which base and height are used, bh must be THE SAME in each case.

Triangle ADY:
If DY is the base, then DA is the corresponding height.
If AY is the base, let h = the corresponding height (the perpendicular from D to DY).
Since the product must be the same in each case, we get:
(DY)(DA) = (AY)(h)
(√6)(√6) = (√12)(h)
h = √36/√12 = √3.
Last edited by GMATGuruNY on Fri Sep 20, 2013 6:07 am, edited 1 time in total.
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by mevicks » Fri Sep 20, 2013 6:07 am
theCodeToGMAT wrote:
Mevrick, there's one rule; The rule says:

Image

where H is the Perpendicular on Hypotenuse
Is this rule valid for all types of Right angled triangles or just Isosceles Right angled triangle ones ? Seems very handy, especially during crunch time :twisted:

Edit: Oh! got it. (1/2)*b*h = (1/2)*b*h = Area
i.e. bh = bh and thus H= ab/c
No matter what the base and height is. So it should work for all right angled triangles :D

Thanks to Mitch n Rahul for the explainations!
Last edited by mevicks on Fri Sep 20, 2013 6:15 am, edited 1 time in total.
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by theCodeToGMAT » Fri Sep 20, 2013 6:11 am
mevicks wrote:
Is this rule valid for all types of Right angled triangles or just Isosceles Right angled triangle ones ? Seems very handy, especially during crunch time :twisted:
Yes, I have solved many questions considering this rule. There is no such constraint on the values of a,b, and c; they may vary
Yeah it's a very Handy rule!!!.
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by vinay1983 » Fri Sep 20, 2013 8:24 am
mevicks wrote:Image

In the figure above, ABCD is a square, and the two diagonal lines divide it into three regions of equal area. If AB = 3, what is the length of w, the perpendicular distance between the two diagonal lines?

A) 3√2 - 2√3
B) 3√2 - √6
C) √2
D) Image
E) 2√3 - √6

Time yourself when solving this, and do post your time ! In exam conditions it took me 3.2 Minutes and then I had to guess and move on !
After reviewing the problem in untimed conditions I found out that there are a few "very easy under 2 minutes" solutions available.

[spoiler]OA A : 3√2 - 2√3[/spoiler]
I feel this is like this:

The length of the diagonal of a square is d = √2

So d = 3√2

Since it is 45-45-90 triangle on both sides, the third side or hypotenuse has to √2, so 2 triangles, hence 2√2. The length w = 3√2-2√2.

If I am wrong, where am i wrong?
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by theCodeToGMAT » Fri Sep 20, 2013 8:53 am
vinay1983 wrote:
I feel this is like this:

The length of the diagonal of a square is d = √2

So d = 3√2

Since it is 45-45-90 triangle on both sides, the third side or hypotenuse has to √2, so 2 triangles, hence 2√2. The length w = 3√2-2√2.

If I am wrong, where am i wrong?
Vinay, applied a rule with a wrong assumption:

In a 45-90-45 triangle,
two non-hypotenuse sides are a & a and not 1 & 1.
So, the Hypotenuse you calculated has to be √2a and not √2.

You need to calculate the measurement of "a"; refer solutions provided by Mitch and me.
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by vinay1983 » Fri Sep 20, 2013 9:29 pm
theCodeToGMAT wrote:
vinay1983 wrote:
I feel this is like this:

The length of the diagonal of a square is d = √2

So d = 3√2

Since it is 45-45-90 triangle on both sides, the third side or hypotenuse has to √2, so 2 triangles, hence 2√2. The length w = 3√2-2√2.

If I am wrong, where am i wrong?
Vinay, applied a rule with a wrong assumption:

In a 45-90-45 triangle,
two non-hypotenuse sides are a & a and not 1 & 1.
So, the Hypotenuse you calculated has to be √2a and not √2.

You need to calculate the measurement of "a"; refer solutions provided by Mitch and me.
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by lunarpower » Sat Sep 21, 2013 6:25 am
I think I wrote this problem, but I honestly can't remember.
theCodeToGMAT wrote:Mevrick, there's one rule; The rule says:

Image

where H is the Perpendicular on Hypotenuse
This is true, but you don't need this much ammunition here.

In that picture for this problem, the yellow thing divides XDY into two 45º-45º-90º triangles. Therefore, the yellow thing is just DX or DY divided by √2.
You already know DX = DY = √6, so the yellow thing is √3.
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by lunarpower » Sat Sep 21, 2013 6:31 am
Also, you can solve this problem by just looking at the picture, estimating the length, and estimating the answer choices.

The picture is accurate, so we can estimate visually. Given that AB = 3, a quick glance makes it clear that w is 1 or less.
mevicks wrote:A) 3√2 - 2√3
Approximately 3(1.4) - 2(1.7).
Approximately 4.2 - 3.4.
Approximately 0.8.
Pretty good.
B) 3√2 - √6
Approximately 4.2 - (something about halfway between 1 and 2).
Whoa! Way way way too big.
C) √2
About 1.4.
Way too big.
(D) 3√2/2
Approximately one and a half times 1.4.
More than 2!
Way way way too big.
E) 2√3 - √6
Approximately 2(1.7) - (something about halfway between 1 and 2).
Approximately 3.4 - (something about halfway between 1 and 2).
Almost 2!
Way way way too big.

A is the only choice that's even halfway reasonable. If you figure out about how big the other four choices are, they're all ridiculous.
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by mevicks » Sat Sep 21, 2013 7:14 pm
@ Ron : Yep, the source is a Manhattan CAT Exam. Wanted some more alternate and faster explanations so posted it here.

Thanks & Regards,
Vivek
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by GMATinsight » Thu Oct 22, 2015 6:19 am
mevicks wrote:Image

In the figure above, ABCD is a square, and the two diagonal lines divide it into three regions of equal area. If AB = 3, what is the length of w, the perpendicular distance between the two diagonal lines?

A) 3√2 - 2√3
B) 3√2 - √6
C) √2
D) Image
E) 2√3 - √6

Time yourself when solving this, and do post your time ! In exam conditions it took me 3.2 Minutes and then I had to guess and move on !
After reviewing the problem in untimed conditions I found out that there are a few "very easy under 2 minutes" solutions available.

[spoiler]OA A : 3√2 - 2√3[/spoiler]
Answer: Option A
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