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# DS : TRIANGLE (m09q07)

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Re: DS : TRIANGLE (m09q07) [#permalink]

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22 Feb 2010, 12:46
1.The area S = a (proven in the above posts)..
tanA=tanC=BO/AO=a^2/(1/a)=a^3.
If angle ABC<90 then A =(180-ABC)/2 > 45, -> tanA>1 -> a^3>1 -> Area S=a >1.
2. Perimeter P = (2/a)(1 + sqrt(a^6+1) > 4/a.
1 + sqrt(a^6+1) > 2
sqrt(a^6+1) > 1 for any a not equal to 0.
Rhe Stmnt 2 alone is not suff.
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Re: DS : TRIANGLE (m09q07) [#permalink]

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23 Feb 2010, 03:56
Bunuel,
For (1), how did you deduce that ABC is an isosceles right triangle ? As angle ABC is less than 90 deg, it is only isosceles at the most. We need not have 45-45 each at the lower angles as well. It could be a 50-50-80 (adding to 180) triangle but never a right triangle. Or am I missing something ??

Also, what do you think of AB^2 + BC^2 > AC^2 ??? I think the rule is right but never quite heard of it before.
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Re: DS : TRIANGLE (m09q07) [#permalink]

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23 Feb 2010, 07:40
SudiptoGmat wrote:
deepakdewani wrote:

How does angle ABC < 90 deg. lead to the conclusion "AB^2 + BC^2 > AC^2"?

Quote:
Its rule. Just memorise it.

While I agree that eventually remebering this rule for GMAT will be the best bet, I am hoping that you can provide the underlying rationale/logic for this rule. Haven't quite come across this rule in the strategy guides...though i am sure this rule can be quite handy in geometry questions involving some variation of inequalities.
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Re: DS : TRIANGLE (m09q07) [#permalink]

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23 Feb 2010, 08:30
This rule is called cosine theorem: for any triangle ABC
$$AC^2=AB^2+BC^2-2AB*BC*cos\angle ABC$$.
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Re: DS : TRIANGLE (m09q07) [#permalink]

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23 Feb 2010, 09:52
Expert's post
kaptain wrote:
Bunuel,
For (1), how did you deduce that ABC is an isosceles right triangle ? As angle ABC is less than 90 deg, it is only isosceles at the most. We need not have 45-45 each at the lower angles as well. It could be a 50-50-80 (adding to 180) triangle but never a right triangle. Or am I missing something ??

Also, what do you think of AB^2 + BC^2 > AC^2 ??? I think the rule is right but never quite heard of it before.

There was an assumption missing: "assume $$\angle ABC=90^\circ$$", already edited.

As for $$AB^2 + BC^2 > AC^2$$: if ABC=90 then we would have AB^2+BC^2=AC^2, but if we decrease angle ABC, value of AC will decrease too and we'll get $$AB^2 + BC^2 > AC^2$$.

Hope it's clear.
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Re: DS : TRIANGLE (m09q07) [#permalink]

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01 Mar 2010, 17:23
hello everyone,

How does angle ABC < 90 deg. lead to the conclusion "AB^2 + BC^2 > AC^2"?

doesn't this appear to be an application of the third side rule of a + b > c but a - b < c so by looking at the graph u can tell how this info was deduced.
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Re: DS : TRIANGLE (m09q07) [#permalink]

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16 Oct 2010, 14:07
Bunuel, how is it deduced as isosceles triangle?
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Re: DS : TRIANGLE (m09q07) [#permalink]

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02 Dec 2010, 12:00
Bunuel,

Great explanation - I am not sure about your conclusion "Hence a>1 is not true as when we decrease angle ABC, a will decrease as well and will become less than 1"

I think this may lead to "As the angle ABC decreases, since ht=a^2 when base = 2a (ht/base ratio = a/2), the base will become broader => a will increase

=> This triangle will have area = 1 for ABC = 90*

=> This triangle will have area > 1 for ABC < 90*

=> This triangle will have area < 1 for ABC > 90* "

Thx, JS
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Re: DS : TRIANGLE (m09q07) [#permalink]

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02 Dec 2010, 15:40
ooops, and my typo 2a in place of 2/a, led to a/2 instead of a^3/2

thank you so much for responding, you are one of the most brilliant minds I've seen in the GMAT math domain and it's always a pleasure to view your problem solving strategies!

JS.
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Re: DS : TRIANGLE (m09q07) [#permalink]

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25 Feb 2011, 08:13
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Re: DS : TRIANGLE (m09q07) [#permalink]

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25 Feb 2011, 10:30
Bunuel
cos(x) when x < 90 is positive and when x > 90 is negative.

When x < 90
ab^2 + bc^2 - (something) = ac^2
or
ab^2 + bc^2 = (something) + ac^2
Hence ab^2 + bc^2 > ac^2

When x > 90
ab^2 + bc^2 + (something) = ac^2
or ab^2 + bc^2 < ac^2

nvgroshar wrote:
This rule is called cosine theorem: for any triangle ABC
$$AC^2=AB^2+BC^2-2AB*BC*cos\angle ABC$$.
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Re: DS : TRIANGLE (m09q07) [#permalink]

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27 Feb 2011, 22:58

How does angle ABC < 90 deg. lead to the conclusion "AB^2 + BC^2 > AC^2"?

[quote]Its rule. Just memorise it.
While I agree that eventually remebering this rule for GMAT will be the best bet, I am hoping that you can

I think, This can be explained by thinking like this, if ABC is 90 degree then AB^2 + BC^2 = AC^2. Now, since side of the triangle is proprtional to the angle opposite to the side- we know that angle ABC is less than 90, so the side opposite to that will be smaller than the hypotenues of the right angle triangle. By this, we can write that AB^2 + BC^2 > AC^2.

Hope it helps!!!
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Re: DS : TRIANGLE (m09q07) [#permalink]

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02 May 2011, 14:22
themanwithaplan wrote:
hello everyone,

How does angle ABC < 90 deg. lead to the conclusion "AB^2 + BC^2 > AC^2"?

doesn't this appear to be an application of the third side rule of a + b > c but a - b < c so by looking at the graph u can tell how this info was deduced.

(realise this is an old question, but for anyone wondering why....)

setting ABC = 90 deg, you get a right angled isos triangle.... pythagoras theorem gives you a^2 + b^2 = c^2
(where c^2 = hypotenuse. and a and b are the other two sides)

logically, if the angle is less than 90 degrees and the triangles is isosceles, side C *must* be smaller than a^2 + b^2....

here they've just used 'AB' and 'BC' to indicate sides 'a' and 'b' as they appear in the theorem, and AC would be the hypotenuse if the triangle was a right triangle
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Re: DS : TRIANGLE (m09q07) [#permalink]

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06 Jun 2011, 08:33
Case 1:
Consider angle ABC=90 , in that case BCA and BAC angles will equal 45 deg.

now applying tan BCA= tan 45 = BO/OC consider O as the origin.

which implies BO/OC = a^3 = 1 or a=1.

Now since its given that angle ABC is less than 90 deg that means angle BAC and BCA can only exceed the value of 45 deg.

And tan of any value in excess of 45 is always more than 1.

That means for angle ABC < 90 we have BO/OC=a^3 > 1 => a > 1 --- sufficient

Case 2:
Permiter is > 4/a
=>AB+BC+CA > 4/a
x+x+ 2/a > 4/a
=> x > 1/a that is BC or BA > CA which is always the case (hypotnuse is alws greater) .Thsi option does nt tell u whther BCA and BAC angles are greater than 45 degree or not.Hence no way to find whther a > 1 or not. -----insuff

Hence ans-A
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Re: DS : TRIANGLE (m09q07) [#permalink]

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20 Dec 2011, 11:33
can anyone explain how they deduced from the illustration that the triangle was an isosceles? The diagram doesn't indicate that AB and AC are equal, so why couldn't it be a 30-60-90 triangle and the illustration is not to scale? is it because from the origin the distance is 1/a both ways, deducing that from that point reaching a2 indicates that ab and bc must be equal?
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Re: DS : TRIANGLE (m09q07) [#permalink]

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29 Feb 2012, 14:49
Tough question for me. Don't think I could do it in under 2 minutes.
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Re: DS : TRIANGLE (m09q07) [#permalink]

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01 Mar 2012, 07:04

Given that ABC is an equilateral triangle (will tell you why it is an equilateral triangle at the end) with side length of 2/a and altitude of length square(a), so the area will be $$\frac{1}{2}$$*$$\frac{2}{a}$$*square(a) = a so the question is whether a>1? And we cannot determine this with the information given in the question so let's look at the given 2 statements whether they provide any extra information:

stmt1 is redundant information - we already know that all angles of an equilateral triangle are of 60 degree
stmt2 is redundant information - length of one side is given (2/a), so perimeter will be (3 * $$\frac{2}{a}$$) = 6/a

So both the statements are not providing any extra information so we cannot determine whether a>1?

Why the given triangle is an equilateral triangle

Triangle property: if altitude, median and perpendicular bisector of a triangle are same, then the triangle is an equilateral triangle.

amitdgr wrote:
On the picture below, is the area of the triangle $$ABC$$ greater than 1?

1. $$\angle ABC < 90^\circ$$
2. Perimeter of the triangle $$ABC$$ is greater than $$\frac{4}{a}$$

[Reveal] Spoiler: OA
A
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Re: DS : TRIANGLE (m09q07) [#permalink]

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01 Mar 2012, 07:16
Expert's post
yatendragoel wrote:
Triangle property: if altitude, median and perpendicular bisector of a triangle are same, then the triangle is an equilateral triangle.

That's not true. If the altitude of a triangle is also the median, then the triangle is not always equilateral, it's always isosceles.

By the way OA for this question is A, not E. OA is given under the spoiler in the initial post.
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Re: DS : TRIANGLE (m09q07) [#permalink]

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03 Mar 2012, 08:42
Tough one. Is there a way by which it can be solved easily
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Re: DS : TRIANGLE (m09q07) [#permalink]

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25 Oct 2012, 14:33
I did this problem differently and was curious if it is a feasible method.

Since ABC is isoceles around the origin, I split ABC into two right triangles and dealt with just one side at a time. Using O as the origin, I drew triangle BOC with sides 1/a and a^2 and angle BOC = 90 deg.

(1) ABC < 90 deg means OBC < 45 deg. Assuming OBC = 45 would give us our limit on variable a. If OBC = 45 deg we would have a 45-45-90 triangle with a^2 (side OB) = 1/a (side OC). Since OBC < 45 deg, OCB must be greater than 45 deg and a^2 (side opposite OCB) must be greater than 1/a (side opposite OCB). Quickly trying different numbers in inequality a^2 > 1/a I found that a>1. Sufficient.

(2) Perimeter > 4/a means that the sides of the triangle are 1/a, 1/a and 2/a (bottom side). Taking just one side and making a right triangle and the Pythagorean theorem, I got (1/a)^2 + (a^2)^2 = (1/a)^2 as the minimum limit. This reduces to a^4>0. Since the perimeter of the big triangle is greater than 4/a, the hypotenuse of my "mini" right triangle is greater than 1/a. Therefore a must be greater than 0. This is however still insufficient.

Thoughts? Please let me know if there are any holes in my reasoning.

Thanks,
HImba88

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Re: DS : TRIANGLE (m09q07)   [#permalink] 25 Oct 2012, 14:33

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# DS : TRIANGLE (m09q07)

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