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The circle above circumscribes square ABCD. What is the valu

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The circle above circumscribes square ABCD. What is the valu [#permalink] New post 07 Sep 2010, 20:57
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A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

32% (03:41) correct 68% (02:24) wrong based on 62 sessions
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The circle above circumscribes square ABCD. What is the value of w°+x°+y°+z°?

A. 180°

B. 135°

C. 120°

D. 100°

E. 90°
[Reveal] Spoiler: OA

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Re: PS - Inscribed angles [#permalink] New post 07 Sep 2010, 22:07
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rafi wrote:
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The circle above circumscribes square ABCD. What is the value of w°+x°+y°+z°?

A. 180°

B. 135°

C. 120°

D. 100°

E. 90°

How would you approach such a question? is a waste of time to try and answer it? or is there a quick shortcut?
Thanks :!:


Wow. Really good one :-D -- finding it hard to come up some sort of a approach to this problem. Just curious -- what is the source of this question.
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Re: PS - Inscribed angles [#permalink] New post 07 Sep 2010, 22:09
It's from a Master GMAT test
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Re: PS - Inscribed angles [#permalink] New post 07 Sep 2010, 22:11
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This is quite a hard question -

I tried to set up a few equations and substitute angles out to hope to get the total value but I onl got x-w-y+z=0 which means x+y+w+z=2(x+z) =2(w+y).

If you imagine the center of the square, called it O. and imagine where Y hits the circle, call it E. BOE=2X and BOC=2Z (Angles that share the same arc are of same value and one with the center of cicle as the top point is always twice as the one with an arbitraty point on the circle) and BOC=90 from the square therefore, w+x+y+z is 90.


It took me more than 3 minutes to solve this and definitely is not recommended on the test. I would have just guessed something is a divident or multiples of 90, given the circle and the square. A, C or E are in the run. If you also look at the image, I wouls have guessed either 90 or 120. The rest of it is your luck under pressure.

Hope this helps.

rafi wrote:
Image

The circle above circumscribes square ABCD. What is the value of w°+x°+y°+z°?

A. 180°

B. 135°

C. 120°

D. 100°

E. 90°

How would you approach such a question? is a waste of time to try and answer it? or is there a quick shortcut?
Thanks :!:

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Re: PS - Inscribed angles [#permalink] New post 07 Sep 2010, 22:25
Thanks for the answer. But I still don't get it. BOC is not on the same arc as angle Z. If I used your E (where y hits the circle) I'd say EOC is on the same arc. Anyhow I still don't understand how you went from this data to the conclusion that the 4 angles equals 90 degrees.
Thanks again!
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Re: PS - Inscribed angles [#permalink] New post 07 Sep 2010, 23:38
Babyif, I think you are naming the variables incorrectly. As Rafi pointed out...

Last edited by hemanthp on 08 Sep 2010, 01:53, edited 1 time in total.
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Re: PS - Inscribed angles [#permalink] New post 08 Sep 2010, 01:48
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E it is.
how:
We all agree that X+Z = W+Y (draw parallel lines at X and Y and you see that the alternate angles match up giving you this condition)

From the figure you can see that the angles subtended by at the center by those arcs that subtend X and Z at the circumference will be 2X and 2Z. Which is the same angle subtended by the chord BC at the center which is 90 for a square.

Similarly for W+Y.

so 2(X+Z) = 90 or X+Z = 45. And therefore X+Z+W+Y = 90.

Clear as spring water. Good question.
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Re: PS - Inscribed angles [#permalink] New post 08 Sep 2010, 02:38
Excellent answer! Kudos!
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Re: PS - Inscribed angles [#permalink] New post 08 Sep 2010, 06:38
Hi guys:
I might have named the angles wrong but on the way to work this morning - I thought about quick way to solve this problem without drawing parallel lines or set up equations. Using algebra to solve geometry problems usually takes longer, nevertheless should arrive the same conclusion.

Ready? I'd love to draw it out as Heman did but I have no scanner in hand now.

I am going to call the left unnamed point in on the circle E ( between A and D). And similarly the right one F. Center of the circle which is also center of the square is O. So connect OA, OE and OD. You will get three lines similar to what Rafi drew. BOF=2x,COF=2z, AOE=2W and EOD=2Y. Because BOF and X share the same length of arc and similarly the other three pairs. It's easy to know that BOF+COF=BOC=90 from the property of square ( or circle). And AOE+EOD=AOD=90; 2x+2Z+2W+2Y= 90+90=180. Thus answer is 90.

This should be how the test writer wanted us to solve the problem. Other than lucky guesses, this is the fastest way I can think of. And it did not come to me immediately. Hope this explanation is easier to read. Very hard to explain a geormetry problem by words only. hehe.

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Re: PS - Inscribed angles [#permalink] New post 08 Sep 2010, 08:16
I think Babyif got it right the first time. And I didn't use a scanner dude. It is PAINT :).
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Re: PS - Inscribed angles [#permalink] New post 09 Sep 2010, 01:18
Hii,

I think there is a simpler way to get this,

See ABCD is a square so we immediately know all the four arcs are same adn corresponding angles formed by square are 90 degree Now if we assume a diagonal BD the angle BDC is 45 ( In square diagonal bisects opposite angles) and this angle subtends the same arc as X & Z since same arc is subtended the sum X + Z = 45

The same logic works for the other two as well and the sum is 90.

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Re: PS - Inscribed angles [#permalink] New post 09 Sep 2010, 10:15
@vivek: Very good use of concepts...
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Re: PS - Inscribed angles [#permalink] New post 27 Dec 2010, 02:57
@Bunuel, Can you help with this
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Re: PS - Inscribed angles [#permalink] New post 27 Dec 2010, 06:14
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hirendhanak wrote:
@Bunuel, Can you help with this


Image
The circle above circumscribes square ABCD. What is the value of w°+x°+y°+z°?
A. 180°
B. 135°
C. 120°
D. 100°
E. 90°

The Central Angle Theorem states that the measure of inscribed angle is always half the measure of the central angle.

So the central angle which subtends the minor arc BC equals to twice the angle x+z (as inscribed angle x plus inscribed angle z subtend the same minor arc BC). Now, minor arc BC is 1/4 of the circumference hence the central angle which subtends it equals to 360/4=90 --> 90=2(x+z) --> x+z=45;

Similarly 90=2(w+y) (for minor arc AD) --> w+y=45;

x+z+w+y=90.

Answer: E.

For more on circles and angles check: math-circles-87957.html

Hope it helps.
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Re: PS - Inscribed angles [#permalink] New post 28 Dec 2010, 03:10
Hi,
May be it sounds funny, but by observing the diagram i found it out that 90degree angle is broken into different acute angles w, x, y, z. So if we observe angle ABC = 90, it is broken into w, x, y, z in the order. And i got the correct OA.

Now by seeing so much explanations from everyone, i am confused whether my first observation was correct or not ?

Plz help.
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Re: PS - Inscribed angles [#permalink] New post 10 Nov 2013, 18:53
Hemmanthp explained the solution well but couldn't explain as to why 2(x+z)=90. Bunuel did it well to explain it.
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Re: PS - Inscribed angles [#permalink] New post 07 Mar 2014, 02:08
Bunuel wrote:
hirendhanak wrote:
@Bunuel, Can you help with this


Image
The circle above circumscribes square ABCD. What is the value of w°+x°+y°+z°?
A. 180°
B. 135°
C. 120°
D. 100°
E. 90°

The Central Angle Theorem states that the measure of inscribed angle is always half the measure of the central angle.

So the central angle which subtends the minor arc BC equals to twice the angle x+z (as inscribed angle x plus inscribed angle z subtend the same minor arc BC). Now, minor arc BC is 1/4 of the circumference hence the central angle which subtends it equals to 360/4=90 --> 90=2(x+z) --> x+z=45;

Similarly 90=2(w+y) (for minor arc AD) --> w+y=45;

x+z+w+y=90.

Answer: E.

For more on circles and angles check: math-circles-87957.html

Hope it helps.



Can you please elaborate this?
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Re: PS - Inscribed angles [#permalink] New post 07 Mar 2014, 03:06
Expert's post
PareshGmat wrote:
Bunuel wrote:
hirendhanak wrote:
@Bunuel, Can you help with this


Image
The circle above circumscribes square ABCD. What is the value of w°+x°+y°+z°?
A. 180°
B. 135°
C. 120°
D. 100°
E. 90°

The Central Angle Theorem states that the measure of inscribed angle is always half the measure of the central angle.

So the central angle which subtends the minor arc BC equals to twice the angle x+z (as inscribed angle x plus inscribed angle z subtend the same minor arc BC). Now, minor arc BC is 1/4 of the circumference hence the central angle which subtends it equals to 360/4=90 --> 90=2(x+z) --> x+z=45;

Similarly 90=2(w+y) (for minor arc AD) --> w+y=45;

x+z+w+y=90.

Answer: E.

For more on circles and angles check: math-circles-87957.html

Hope it helps.



Can you please elaborate this?


Can you please tell me which part is problematic? Thank you.
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Re: PS - Inscribed angles   [#permalink] 07 Mar 2014, 03:06
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