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Problem Solving for 780+ Aspirants.

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Problem Solving for 780+ Aspirants.

by sureshbala » Fri Feb 06, 2009 11:52 pm
Folks, I have a work experience of more than 6 years as a GMAT Faculty and Content Developer. In this thread I will post questions on a regular basis keeping in mind the standard of the questions that one needs to be comfortable with in order to score top in Quant Section of the exam.

Here are the two simple rules that I want all the members to strictly adhere to..

1. Do not post your questions in this thread. Instead start a new thread or post in the relevant thread.

2. Please give a bit of explanation for the answer that you post here.(Don't simply post the answer)

Here is our first question.


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In the above figure, ABC is a right angled triangle. Taking the hypotenuse AC a square is constructed. If O is the center of the square find angle ABO.
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by DanaJ » Sat Feb 07, 2009 9:02 am
I am pretty curious to see the explanation for this one. IMHO (although, of course, I might be wrong), there is smth missing.
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by shulapa » Sat Feb 07, 2009 10:36 am
This is a tough one. The only thing I could have done under the pressure of "two minutes per question" limit, and under the assumption that we are looking for a numerical answer, is to look at a particular case. As we are not given any information regarding the angles ACB and CAB and a numerical answer will be the same in any ratio between them, we can assume that they are equal. From here it is fairly easy to prove that OB bisects the angle ABC so that ABO is equal to 45 degrees.
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by zizikathegreat » Sat Feb 07, 2009 11:51 am
There is definitely something is missing. Is it DS problem???

It is known that ABC is right triangle, AC hypotenuse. Without knowing the relationship between the sides of the right triangle we cannot determine the measure of OBC. The right triangles such as 90-45-45, or 90-60-30 - will give us different results.

Please share your answer to this question, as it looks incomplete to me. We cannot give an answer simply by assuming that BO bisects ABC ( 90degrees angle) equally. It shouldn't be so easy :) considering the level of difficulty.

Waiting for your answer to this question.
I can beat the GMAT!!!!!
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by sureshbala » Sat Feb 07, 2009 12:36 pm
Folks, this is definitely a problem to be solved (not DS)

I am sure once you know the answer you will find it easy...Just keep trying
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by shulapa » Sat Feb 07, 2009 3:57 pm
Hello all,

I think I have a way to solve this. It came as an inspiration in the middle of the night (as it is almost 2 am in this side of the world) and I really look forward to see your opinions.

Lets first draw two lines from O to the left corners of the square. Now we can determine that the new lines (AO and OC) are equals and the angle between them, AOC, is a right angle.

Now we can also drow a circle with AC as its diameter. Knowing that angles ABC and AOC are right angle we can tell that they are placed on the circle.

Now, because we know that AO and OC are equal we know that the angles in front of them (on the circle) should be equal as well. Therefore, ABO = CBO = 45 degrees.

I urge you to rethink my approach of considering the obvious case in which AB = BC. This method provides the same answer but much quicker.

The use of more advance mathematical topic of limits can show that when BC is taken to a close minimum of zero (and therefore, AB is becoming almost AB) we can reach the same conclusion as well.

Now I hope I can finally go to sleep.

Eyal
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Last edited by shulapa on Sun Feb 08, 2009 9:08 am, edited 1 time in total.
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by sureshbala » Sat Feb 07, 2009 4:51 pm
Dear shulapa,

Congrats for the correct answer(approach as well)

Small correction....

Draw a circle with AC as the diameter and not OC as the diameter. I think that's a typo...please make the necessary correction.
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by ahhock » Sat Feb 07, 2009 7:00 pm
Just because AO = OC doesn't mean ABO = CBO rite? There's an infinite number of ways in which triangle ABC can be a right angled triangle, given all the constraints. The above explanation is correct only if triangle ABC is an isoceles triangle.
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by Ian Stewart » Sat Feb 07, 2009 8:29 pm
The angle does need to be 45, regardless of the dimensions of the triangle. There are likely several solutions to the problem; I looked at it using properties of perpendicular lines in co-ordinate geometry:

-Put point B at (0,0) in the x-y plane, and put A on the y-axis, at (0,a), and C on the x-axis, at (c,0), where a and c are positive;

-The slope of line AC is then -a/c;

-Let D be the vertex of the square which is diagonally opposite C. The line AD is both perpendicular to and equal in length to AC. To go from A to C, we go right c units and down a units. To go the same distance on a perpendicular line, we need to go right a units and up c units (perpendicular slopes are negative reciprocals of each other). So the point D will be a units to the right of A and c units up from A; that is, it will be at (a, a+c);

-the centre of the square is the midpoint of a diagonal of the square; that is, the centre is at the midpoint of C = (c, 0) and D = (a, a+c). The centre is thus at ((a+c)/2, (a+c)/2));

-we see that the centre of the square has an equal x and y co-ordinate - it lies on the line y=x. This line makes an angle of 45 degrees with the y-axis when it crosses through the origin, and that's exactly angle ABO.
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by sureshbala » Sat Feb 07, 2009 11:38 pm
ahhock wrote:Just because AO = OC doesn't mean ABO = CBO rite? There's an infinite number of ways in which triangle ABC can be a right angled triangle, given all the constraints. The above explanation is correct only if triangle ABC is an isoceles triangle.
Dear ahhock, angles made by two equal chords on the same segment of the circle are always equal. So definitely angle ABO = angle CBO.

By the way Ian, the way in which you solved is also good but I think that if one can get to the simple rules of circles and chords, this can be answered quickly.
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by sureshbala » Sun Feb 08, 2009 12:05 am
Folks, let me conclude this question and go to the next one. But before that here is a quick solution for this question.

Image

Since LAOC and LABC are both right. the circle on diameter AC passes through B and O. Since AO and OC are equal chords of this circle they subtend equal angles at the circumference(major segment). So LABO = LCBO = 45 degrees.
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by 4meonly » Sun Feb 08, 2009 12:31 am
Guys, I do not understand how you solved this problem.
Since we have circle ABCO we can take ANY point for B between A and C on the circle and in ALL this cases angle B will be eq. to 90 degrees. In ALL cases AO=CO since they are diagonals. It does not guarantee that BO will bisect AOC.
Therefore, I think that this problem lacks some necessary statement. Otherwise, I miss something.
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by sureshbala » Sun Feb 08, 2009 1:29 am
4meonly wrote:Guys, I do not understand how you solved this problem.
Since we have circle ABCO we can take ANY point for B between A and C on the circle and in ALL this cases angle B will be eq. to 90 degrees. In ALL cases AO=CO since they are diagonals. It does not guarantee that BO will bisect AOC.
Therefore, I think that this problem lacks some necessary statement. Otherwise, I miss something.
Dear 4meonly, look at the bold statement above.

We did not conclude that BO will bisect AOC, instead what we concluded is BO bisects angle ABC. Irrespective of the position of B (between A and C) on the circumference of the circle, since AO and OC are equal chords angle made them at B must be equal. Thus angle ABO = angle OBC

So there's nothing wrong with this question and the data is absolutely sufficient to answer this question
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by 4meonly » Sun Feb 08, 2009 2:04 am
Sorry, sureshbala, I still disagree.
we concluded is BO bisects angle ABC.
I really do not understand how you did it :?
ahhock wrote:Just because AO = OC doesn't mean ABO = CBO rite? There's an infinite number of ways in which triangle ABC can be a right angled triangle, given all the constraints. The above explanation is correct only if triangle ABC is an isoceles triangle.
I absolutely agree with ahhock. The only way to make <ABO = <CBO is to make AB=BC. It is not stated anywhere.
I also can prove that data is insuficient by counterexamples.
You can find them on the picture below. Points B1 and B3 both satisfy the statement of the problem. Who can prove that <AOB1 = <COB1 = 45 degrees?
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by sureshbala » Sun Feb 08, 2009 2:41 am
4meonly wrote:Sorry, sureshbala, I still disagree.
we concluded is BO bisects angle ABC.
I really do not understand how you did it :?
ahhock wrote:Just because AO = OC doesn't mean ABO = CBO rite? There's an infinite number of ways in which triangle ABC can be a right angled triangle, given all the constraints. The above explanation is correct only if triangle ABC is an isoceles triangle.
I absolutely agree with ahhock. The only way to make <ABO = <CBO is to make AB=BC. It is not stated anywhere.
I also can prove that data is insuficient by counterexamples.
You can find them on the picture below. Points B1 and B3 both satisfy the statement of the problem. Who can prove that <AOB1 = <COB1 = 45 degrees?
Image
Dear friend, again look at the bold statement ...there's an error.

I think you want me to prove that &#8735;AB1O = &#8735;CB1O.

This is pretty simple.

Let &#8735;AB1O = &#952; i.e angle made by the chord AO at B1. Since the angle made by the chord at any point on the major(minor) segment is same, the angle made by this chord AO at C also must be the same &#952;.
Thus &#8735;ACO = &#952;.

Since in the triangle ACO, AO = OC obviously &#8735;ACO = &#8735;OAC
Hence &#8735;OAC = &#952;.

Now &#8735;OAC is nothing but the angle made by the chord OC at A. Applying the same logic that we have done earlier i.e since the angle made by the chord at any point on the major(minor) segment is same, the angle made by this chord OC at B1 also must be &#952;. Thus &#8735;CB1O = &#952;.

Hence &#8735;AB1O = &#8735;CB1O = &#952;.

Thus in a circle, the angles made by two chords of equal length at any point on the major(minor) segment of the circle will be same.

I hope this is clear....
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