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Points A, B, C and D lie on a circle of radius 1. Let x be

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Points A, B, C and D lie on a circle of radius 1. Let x be [#permalink] New post 01 Dec 2009, 22:57
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Points A, B, C and D lie on a circle of radius 1. Let x be the length of arc AB and y the length of arc CD respectively, such that x \lt \pi and y \lt \pi . Is x \gt y ?

1. \angle ADB is acute
2. \angle ADB > \angle CAD

Last edited by study on 02 Dec 2009, 11:15, edited 1 time in total.
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Re: Circle [#permalink] New post 02 Dec 2009, 00:59
study wrote:
Points A, B, C and D lie on a circle of radius 1. Let x be the length of arc AB and y the length of arc CD respectively, such that x \lt \pi and y \lt \pi . Is x \gt y ?

1. \angle ADB is acute
2. \angle ADB > \angle CAD



can you please clarify on the expression in question. not able to make out :|
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Re: Circle [#permalink] New post 03 Dec 2009, 01:10
I think it is B
What is the OA?

Given all points are on the arc, angle ADB>angle CAD corresponds to arcs x and y. x>y
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Re: Circle [#permalink] New post 03 Dec 2009, 14:30
Is it B?

A. The fact that angle ADB is acute doesn't indicate anything about arc CD. Therefore it's insufficient
B. ADB corresponds to x and CAD correspons to y. Therefore if ADB>CAD then x is greater than y.
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Re: Circle [#permalink] New post 05 Dec 2009, 12:10
looks clear B to me

inscribed angle describes the arc proportional to 180 degrees
So, ADB > CAD means x >y

what is OA?
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Re: Points A, B, C and D lie on a circle of radius 1. Let x be [#permalink] New post 27 Sep 2012, 04:45
Bunuel,
OA for this one says C. Can you please clarify why it is so?

Thanks
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Re: Points A, B, C and D lie on a circle of radius 1. Let x be [#permalink] New post 27 Sep 2012, 11:18
I also fell for trap B, then realized.

I think answer must be C only, because order of the points on the circle were not mentioned, only if you consider that ADB<90 then only we know B is between A and D.

Correct me If any other explanation is possible.

Thanks,
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Re: Points A, B, C and D lie on a circle of radius 1. Let x be [#permalink] New post 30 Sep 2012, 17:36
If the angle subtended by the arc is greater, the length of the arc has to be greater. I am not sure why C) is OA....help?
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Re: Points A, B, C and D lie on a circle of radius 1. Let x be [#permalink] New post 01 Oct 2012, 03:03
voodoochild wrote:
If the angle subtended by the arc is greater, the length of the arc has to be greater. I am not sure why C) is OA....help?


(2) alone is not sufficient.
Remember the condition in (1) and check the case when that condition doesn't hold.
In my attached drawing, in the isosceles trapezoid, x=y.
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Re: Points A, B, C and D lie on a circle of radius 1. Let x be [#permalink] New post 06 Oct 2012, 01:24
study wrote:
Points A, B, C and D lie on a circle of radius 1. Let x be the length of arc AB and y the length of arc CD respectively, such that x \lt \pi and y \lt \pi . Is x \gt y ?

1. \angle ADB is acute
2. \angle ADB > \angle CAD



Well We can make a comparison between the 2 arcs only if its given explicitly.

Statement 1 says Angle ADB is acute and we cannot say here if x is greater than y.
Statement 2 says one angle is greater than the other. So this must be sufficient

Is there an explanation for Y it is C.
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Re: Points A, B, C and D lie on a circle of radius 1. Let x be [#permalink] New post 07 Oct 2012, 22:09
OA is B for me . Kindly Confirm Bunuel.

cheers :-D
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Re: Points A, B, C and D lie on a circle of radius 1. Let x be [#permalink] New post 08 Oct 2012, 01:25
154238 wrote:
OA is B for me . Kindly Confirm Bunuel.

cheers :-D


No, B is not the answer. See my previous post above points-a-b-c-and-d-lie-on-a-circle-of-radius-1-let-x-be-87495.html#p1126726
In the drawing, an isosceles trapezoid is depicted, and obviously x=y.
For example, by moving point B, you can increase or decrease the arc AB, while arc DC stays the same. So, you can have both x>y and x<y, after you already had x=y.
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Re: Points A, B, C and D lie on a circle of radius 1. Let x be [#permalink] New post 08 Oct 2012, 03:14
rajathpanta wrote:
study wrote:
Points A, B, C and D lie on a circle of radius 1. Let x be the length of arc AB and y the length of arc CD respectively, such that x \lt \pi and y \lt \pi . Is x \gt y ?

1. \angle ADB is acute
2. \angle ADB > \angle CAD



Well We can make a comparison between the 2 arcs only if its given explicitly.

Statement 1 says Angle ADB is acute and we cannot say here if x is greater than y.
Statement 2 says one angle is greater than the other. So this must be sufficient

Is there an explanation for Y it is C.


The answer is not B, (2) alone is not sufficient. See my previous post points-a-b-c-and-d-lie-on-a-circle-of-radius-1-let-x-be-87495.html#p1129047

Why the answer is C?

Any two points on the circumference of the circle define two arcs. Only if the two points are the endpoints of a diameter, the two arcs have equal measure. Otherwise, one of the arcs is less than the other. In our case, the circumference of the circle is 2\pi, therefore x and y being less than \pi are both the shorter arcs defined by A,B and C,D respectively.

If we don't know that \angle{ADB} is acute, we don't know whether point D is on the shorter or the longer arc defined by A and B. From (1), we have that \angle{ADB} is acute. In the attached drawing, D can be only above the chord AB. Therefore to \angle{ADB} corresponds the shorter arc x.
From (2) we have that \angle{ADB}>\angle{CAD}, meaning that \angle{CAD} is also acute. Therefore the corresponding arc to \angle{CAD} is the shorter arc y. In the attached drawing C can be only on the right of the diameter DC_1.
Using (2), now we can conclude that x>y.
So, (1) and (2) together is sufficient.

Answer C.
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Re: Points A, B, C and D lie on a circle of radius 1. Let x be [#permalink] New post 09 Oct 2012, 01:28
EvaJager wrote:
rajathpanta wrote:
study wrote:
Points A, B, C and D lie on a circle of radius 1. Let x be the length of arc AB and y the length of arc CD respectively, such that x \lt \pi and y \lt \pi . Is x \gt y ?

1. \angle ADB is acute
2. \angle ADB > \angle CAD



Well We can make a comparison between the 2 arcs only if its given explicitly.

Statement 1 says Angle ADB is acute and we cannot say here if x is greater than y.
Statement 2 says one angle is greater than the other. So this must be sufficient

Is there an explanation for Y it is C.


The answer is not B, (2) alone is not sufficient. See my previous post points-a-b-c-and-d-lie-on-a-circle-of-radius-1-let-x-be-87495.html#p1129047

Why the answer is C?

Any two points on the circumference of the circle define two arcs. Only if the two points are the endpoints of a diameter, the two arcs have equal measure. Otherwise, one of the arcs is less than the other. In our case, the circumference of the circle is 2\pi, therefore x and y being less than \pi are both the shorter arcs defined by A,B and C,D respectively.

If we don't know that \angle{ADB} is acute, we don't know whether point D is on the shorter or the longer arc defined by A and B. From (1), we have that \angle{ADB} is acute. In the attached drawing, D can be only above the chord AB. Therefore to \angle{ADB} corresponds the shorter arc x.
From (2) we have that \angle{ADB}>\angle{CAD}, meaning that \angle{CAD} is also acute. Therefore the corresponding arc to \angle{CAD} is the shorter arc y. In the attached drawing C can be only on the right of the diameter DC_1.
Using (2), now we can conclude that x>y.
So, (1) and (2) together is sufficient.

Answer C.


Hi Eva,

we have that \angle{ADB}>\angle{CAD}, how can you conclude that that [m]\angle{CAD} is also acute.
Let say you make a square inside the circle
a---------d
| |
| |
c---------b

now, if \angle{ADB}>\angle{CAD}, this doesnt mean that angle{CAD} is acute. Right ?

Its still not clear to me, why B is wrong ?

Thanks
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Re: Points A, B, C and D lie on a circle of radius 1. Let x be [#permalink] New post 09 Oct 2012, 02:03
EvaJager wrote:
rajathpanta wrote:
study wrote:
Points A, B, C and D lie on a circle of radius 1. Let x be the length of arc AB and y the length of arc CD respectively, such that x \lt \pi and y \lt \pi . Is x \gt y ?

1. \angle ADB is acute
2. \angle ADB > \angle CAD



The answer is not B, (2) alone is not sufficient. See my previous post points-a-b-c-and-d-lie-on-a-circle-of-radius-1-let-x-be-87495.html#p1129047
Why the answer is C?

Any two points on the circumference of the circle define two arcs. Only if the two points are the endpoints of a diameter, the two arcs have equal measure. Otherwise, one of the arcs is less than the other. In our case, the circumference of the circle is 2\pi, therefore x and y being less than \pi are both the shorter arcs defined by A,B and C,D respectively.

If we don't know that \angle{ADB} is acute, we don't know whether point D is on the shorter or the longer arc defined by A and B. From (1), we have that \angle{ADB} is acute. In the attached drawing, D can be only above the chord AB. Therefore to \angle{ADB} corresponds the shorter arc x.
From (2) we have that \angle{ADB}>\angle{CAD}, meaning that \angle{CAD} is also acute. Therefore the corresponding arc to \angle{CAD} is the shorter arc y. In the attached drawing C can be only on the right of the diameter DC_1.
Using (2), now we can conclude that x>y.
So, (1) and (2) together is sufficient.

Answer C.


Hi Eva,

we have that \angle{ADB}>\angle{CAD}, how can you conclude that that [m]\angle{CAD} is also acute.
Let say you make a square inside the circle
a---------d
| |
| |
c---------b

now, if \angle{ADB}>\angle{CAD}, this doesnt mean that angle{CAD} is acute. Right ?

Here, in fact all angles are equal and are right angles. This cannot be the situation in the present question. In addition, if both angles ADB and CAD are right angles, then the inequality \angle{ADB}>\angle{CAD} doesn't hold!
See below.


Its still not clear to me, why B is wrong ?

Thanks



For why B is not sufficient see my previous post points-a-b-c-and-d-lie-on-a-circle-of-radius-1-let-x-be-87495.html#p1129047

When taking (1) and (2) together, from (1) we have that \angle{ADB} is acute, meaning its measure is less than 90 degrees. Then using (2) we get that \angle{CAD}<\angle{ADB}<90^o, so necessarily \angle{CAD} is also acute.
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Re: Points A, B, C and D lie on a circle of radius 1. Let x be   [#permalink] 09 Oct 2012, 02:03
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