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In the figure above, the circles with centers A and B are tangent to [#permalink]
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29 Nov 2017, 23:03
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In the figure above, the circles with centers A and B are tangent to [#permalink]
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29 Nov 2017, 23:46
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Bunuel I have a doubt. Without knowing the angle between the line AB and EF how to calculate the distance between ED? Please help. Please refer the attached figure. For each case the length will be different. Sent from my iPhone using GMAT Club Forum
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In the figure above, the circles with centers A and B are tangent to [#permalink]
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30 Nov 2017, 02:37
Hi, I tried an overly simplistic approach, and got the answer E. It took me 10 seconds after trying all the theories I knew. Basically EF consists of 3 radi's plus something in the middle , so it must be 3√2 + 1. ( assuming the line opposite the radius is parallel to the radius.
Many assumptions taken, still not a concrete answer.



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Re: In the figure above, the circles with centers A and B are tangent to [#permalink]
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21 Dec 2017, 09:48
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septwibowo wrote: Bunuel wrote: In the figure above, the circles with centers A and B are tangent to each other at C, and are tangents to the lines m and n at F and D. If the radius of each of the circles is √2 and m is parallel to n, what is the length of EF? (A) 2√2 + 1 (B) 2√2 + 2 (C) 2√2 + 3 (D) 3√2 (E) 3√2 + 1 Attachment: The attachment 20171130_1000.png is no longer available Bunuel, do we have OE for this answer? Thanks! Hi septwibowoRefer below image. I have joined points G & C where line GC will be common tangent to both the circles Attachment:
geo.jpg [ 19.95 KiB  Viewed 585 times ]
BC & AC will be perpendicular to line GC (radius is perpendicular to the tangent) So Triangles ACG & BCG will be similar (AC=BC=radius; GC is common side, angle GCB=angle GCA=90) => \(\frac{AC}{BC}=\frac{AG}{BG} =>AG=BG\) this means that triangle ABG will be isosceles right angle triangle, so sides will be in the ration \(1:1:\sqrt{2}\) \(AB= \sqrt{2}+\sqrt{2}=2\sqrt{2}\) Therefore \(BG=2\) Hence \(EF= EG+BG+BF\) \(EG=AD=\sqrt{2}\) So \(EF=\sqrt{2}+2+\sqrt{2}=2\sqrt{2}+2\) Option B



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In the figure above, the circles with centers A and B are tangent to [#permalink]
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28 Dec 2017, 01:27
niks18 wrote: septwibowo wrote: Bunuel wrote: In the figure above, the circles with centers A and B are tangent to each other at C, and are tangents to the lines m and n at F and D. If the radius of each of the circles is √2 and m is parallel to n, what is the length of EF? (A) 2√2 + 1 (B) 2√2 + 2 (C) 2√2 + 3 (D) 3√2 (E) 3√2 + 1 Attachment: 20171130_1000.png Bunuel, do we have OE for this answer? Thanks! Hi septwibowoRefer below image. I have joined points G & C where line GC will be common tangent to both the circles Attachment: geo.jpg BC & AC will be perpendicular to line GC (radius is perpendicular to the tangent) So Triangles ACG & BCG will be similar (AC=BC=radius; GC is common side, angle GCB=angle GCA=90) => \(\frac{AC}{BC}=\frac{AG}{BG} =>AG=BG\) this means that triangle ABG will be isosceles right angle triangle, so sides will be in the ration \(1:1:\sqrt{2}\) \(AB= \sqrt{2}+\sqrt{2}=2\sqrt{2}\) Therefore \(BG=2\) Hence \(EF= EG+BG+BF\) \(EG=AD=\sqrt{2}\) So \(EF=\sqrt{2}+2+\sqrt{2}=2\sqrt{2}+2\) Option BHi niks18, How did you conclude that a line tangent to both the circles from point C will meet at point G??



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Re: In the figure above, the circles with centers A and B are tangent to [#permalink]
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28 Dec 2017, 10:33
Krystallized wrote: Hi niks18, How did you conclude that a line tangent to both the circles from point C will meet at point G?? Hi KrystallizedFew thing to note here  1. Point C is the point of tangency of both the circles 2. Radius is perpendicular to the tangent, so AC & BC will be perpendicular to the line passing through point C 3. Point C is also the midpoint of hypotenuse of right triangle AGB, where G is vertex 4. Perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle. You can always join the midpoint of the hypotenuse with the vertex. I do not see any issue here. Is there any specific query you have in mind?



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In the figure above, the circles with centers A and B are tangent to [#permalink]
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28 Dec 2017, 21:00
niks18 wrote: Krystallized wrote: Hi niks18, How did you conclude that a line tangent to both the circles from point C will meet at point G?? Hi KrystallizedFew thing to note here  1. Point C is the point of tangency of both the circles 2. Radius is perpendicular to the tangent, so AC & BC will be perpendicular to the line passing through point C 3. Point C is also the midpoint of hypotenuse of right triangle AGB, where G is vertex 4. Perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle. You can always join the midpoint of the hypotenuse with the vertex. I do not see any issue here. Is there any specific query you have in mind? Hi niks18Thank you for your reply and explaination I totally agree with the points mentioned by you in the above post. My specific query is, how did you decide that a line perpendicular to AB from point C will intersect line BE at G? I think this is possible only if angle ABG and angle BAG are each 45 degree or AG and BG are of same length. Otherwise the line would intersect BE somewhere below point G. I know we can always join the midpoint of hypotenuse with the vertex, but that does not guarantee that the line joining those two points will always be perpendicular to the hypotenuse.



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Re: In the figure above, the circles with centers A and B are tangent to [#permalink]
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29 Dec 2017, 10:24
Krystallized wrote: niks18 wrote: Krystallized wrote: Hi niks18, How did you conclude that a line tangent to both the circles from point C will meet at point G?? Hi KrystallizedFew thing to note here  1. Point C is the point of tangency of both the circles 2. Radius is perpendicular to the tangent, so AC & BC will be perpendicular to the line passing through point C 3. Point C is also the midpoint of hypotenuse of right triangle AGB, where G is vertex 4. Perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle. You can always join the midpoint of the hypotenuse with the vertex. I do not see any issue here. Is there any specific query you have in mind? Hi niks18Thank you for your reply and explaination I totally agree with the points mentioned by you in the above post. My specific query is, how did you decide that a line perpendicular to AB from point C will intersect line BE at G? I think this is possible only if angle ABG and angle BAG are each 45 degree or AG and BG are of same length. Otherwise the line would intersect BE somewhere below point G. I know we can always join the midpoint of hypotenuse with the vertex, but that does not guarantee that the line joining those two points will always be perpendicular to the hypotenuse. Hi KrystallizedI hope you would agree that triangle ABG is a right angled triangle with angle AGB being 90. AB will be the hypotenuse of this triangle with point C as its mid point. Point C is the point of tangency of both the circles so any tangent passing through point C will be perpendicular to AB (AC & BC are radius and radius is perpendicular to the tangent) Next if from point G I draw a perpendicular bisector or simply a bisector of the hypotenuse AB, then Can you tell from which point the line will pass? angle ABG and angle BAG will eventually each be 45 because Perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle and in my solution above it is evident that AG=BG i.e the it is an isosceles right angle triangle. Also note, you cannot squeeze or expand the circles because radius is fixed. You cannot increase the distance between the parallel lines as it is also fixed. So if you are confused that such a situation can also arise at some other position by moving the circles, then that is not possible because in that case either the parallel lines will not be tangent or their distance will have to be increased/decreased. The figure is drawn in such a way that triangle ABG will be right isosceles, there is no other possibility otherwise you will have to alter the distance or the radius. you cannot move the upper circle lower otherwise it will not touch the upper tangent line and you cannot move the circle further high becuase in that case for upper line to be tangent you will have to squeeze i.e reduce the radius of upper circle.




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