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# In the figure above, the circles are centered at O(0, 0) and

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Re: In the figure above, the circles are centered at O(0, 0) and [#permalink]
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Hi There, Can you please elaborate? I kind of figured out that the answer is D, by reasoning that the ratio of sides of the bigger triangle to that of the smaller triangle is 2:1 which led me to x-intercept of 3.33. Am I on the right track?
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Re: In the figure above, the circles are centered at O(0, 0) and [#permalink]
bluecatie1 wrote:
Hi There, Can you please elaborate? I kind of figured out that the answer is D, by reasoning that the ratio of sides of the bigger triangle to that of the smaller triangle is 2:1 which led me to x-intercept of 3.33. Am I on the right track?

Absolutely. Also, once you recognize how to do this from F.S 1, you don't even have to work out anything for the second fact statement. It is all testing the ratio from similar triangles.
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Re: In the figure above, the circles are centered at O(0, 0) and [#permalink]
I still do not understand how to solve this problem. Any help is appreciated.
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Re: In the figure above, the circles are centered at O(0, 0) and [#permalink]
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Bunuel wrote:
blakjedi wrote:
I still do not understand how to solve this problem. Any help is appreciated.

In the figure above, the circles are centered at O(0, 0) and P(10, 0). Line AB is tangent to both circles, at points A and B respectively, and intersects the x-axis at point X. What is the x-coordinate of point X ?

Since AB is tangent to both circles then it's perpendicular to the respective radii to points A and B:
Attachment:
Untitled.png
Also, notice that all three angles in triangles AOX and BPX are equal, which implies that triangles AOX and BPX are similar. This means that the ratio of corresponding sides in these triangles are equal (corresponding sides are opposite equal angles):

(1) The area of the circle centered at point P is 4 times the area of the circle centered at point O --> $$\pi{R^2}=4\pi{r^2}$$ --> $$R=2r$$ --> XP = 2*OX. Since OP = 10 = OX + XP, then OX + 2*OX = 10 --> OX = 10/3 --> the x-coordinate of point X is 10/3. Sufficient.

(2) BX is twice as long as AX. Basically the same here: BX = 2*AX --> --> XP = 2*OX. Since OP = 10 = OX + XP, then OX + 2*OX = 10 --> OX = 10/3 --> the x-coordinate of point X is 10/3. Sufficient.

Hope it's clear.

Many thanks. I doubt I could complete this DS assesment in less than 2:30 minutes under normal circumstances.
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Re: In the figure above, the circles are centered at O(0, 0) and [#permalink]
Bunuel wrote:
blakjedi wrote:
I still do not understand how to solve this problem. Any help is appreciated.

In the figure above, the circles are centered at O(0, 0) and P(10, 0). Line AB is tangent to both circles, at points A and B respectively, and intersects the x-axis at point X. What is the x-coordinate of point X ?

Since AB is tangent to both circles then it's perpendicular to the respective radii to points A and B:
Attachment:
Untitled.png
Also, notice that all three angles in triangles AOX and BPX are equal, which implies that triangles AOX and BPX are similar. This means that the ratio of corresponding sides in these triangles are equal (corresponding sides are opposite equal angles):

(1) The area of the circle centered at point P is 4 times the area of the circle centered at point O --> $$\pi{R^2}=4\pi{r^2}$$ --> $$R=2r$$ --> XP = 2*OX. Since OP = 10 = OX + XP, then OX + 2*OX = 10 --> OX = 10/3 --> the x-coordinate of point X is 10/3. Sufficient.

(2) BX is twice as long as AX. Basically the same here: BX = 2*AX --> --> XP = 2*OX. Since OP = 10 = OX + XP, then OX + 2*OX = 10 --> OX = 10/3 --> the x-coordinate of point X is 10/3. Sufficient.

Hope it's clear.

Hi Bunuel,

Two questions here:

1) How can you tell that "all three angles in triangles AOX and BPX are equal"? I thought that property held when a perpendicular line bisects a hypotenuse and that line is shared by both the triangles. I don't see that here?

2) In statement two, how do you make this leap -- BX = 2*AX --> --> XP = 2*OX?

Thanks
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Re: In the figure above, the circles are centered at O(0, 0) and [#permalink]
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russ9 wrote:
Bunuel wrote:
blakjedi wrote:
I still do not understand how to solve this problem. Any help is appreciated.

In the figure above, the circles are centered at O(0, 0) and P(10, 0). Line AB is tangent to both circles, at points A and B respectively, and intersects the x-axis at point X. What is the x-coordinate of point X ?

Since AB is tangent to both circles then it's perpendicular to the respective radii to points A and B:

Also, notice that all three angles in triangles AOX and BPX are equal, which implies that triangles AOX and BPX are similar. This means that the ratio of corresponding sides in these triangles are equal (corresponding sides are opposite equal angles):

(1) The area of the circle centered at point P is 4 times the area of the circle centered at point O --> $$\pi{R^2}=4\pi{r^2}$$ --> $$R=2r$$ --> XP = 2*OX. Since OP = 10 = OX + XP, then OX + 2*OX = 10 --> OX = 10/3 --> the x-coordinate of point X is 10/3. Sufficient.

(2) BX is twice as long as AX. Basically the same here: BX = 2*AX --> --> XP = 2*OX. Since OP = 10 = OX + XP, then OX + 2*OX = 10 --> OX = 10/3 --> the x-coordinate of point X is 10/3. Sufficient.

Hope it's clear.

Hi Bunuel,

Two questions here:

1) How can you tell that "all three angles in triangles AOX and BPX are equal"? I thought that property held when a perpendicular line bisects a hypotenuse and that line is shared by both the triangles. I don't see that here?

2) In statement two, how do you make this leap -- BX = 2*AX --> --> XP = 2*OX?

Thanks

Triangles are similar if their three angles are identical.

Now, in triangles AOX and BPX: $$\angle{AXO} = \angle{BXP}$$, and $$\angle{OAX} = \angle{PBX}$$, thus their third angles must also be equal: $$\angle{AOX} = \angle{XPB}$$. Therefore triangles AOX and BPX are similar.

As for your second question: since triangles AOX and BPX are similar,then their corresponding sides are all in the same proportion. Thus if BX is twice as long as AX, then XP is twice as long as OX.

Hope it's clear.
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Re: In the figure above, the circles are centered at O(0, 0) and [#permalink]
Bunuel wrote:

Triangles are similar if their three angles are identical.

Now, in triangles AOX and BPX: $$\angle{AXO} = \angle{BXP}$$, and $$\angle{OAX} = \angle{PBX}$$, thus their third angles must also be equal: $$\angle{AOX} = \angle{XPB}$$. Therefore triangles AOX and BPX are similar.

As for your second question: since triangles AOX and BPX are similar,then their corresponding sides are all in the same proportion. Thus if BX is twice as long as AX, then XP is twice as long as OX.

Hope it's clear.

Ahh, got it. I was missing this gap. $$\angle{AXO} = \angle{BXP}$$ Thanks!
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Re: In the figure above, the circles are centered at O(0, 0) and [#permalink]
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This is the beauty of GMAT. This entire horrendous question can be solved just be applying the concept of similar triangles and the formula of internal angle bisector.

If we join the two figures we have two triangles which have 3 angles equal to each other.One is vertically oppsite angle, another one is a right angle and the third one is by default equal. Now if we get the ratio of one side we can immediately get the ratio of the other other sides. and then apply internal bisector formula to devive the answer.
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Re: In the figure above, the circles are centered at O(0, 0) and [#permalink]
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AccipiterQ wrote:
Attachment:
3078_1.png
In the figure above, the circles are centered at O(0, 0) and P(10, 0). Line AB is tangent to both circles, at points A and B respectively, and intersects the x-axis at point X. What is the x-coordinate of point X ?

(1) The area of the circle centered at point P is 4 times the area of the circle centered at point O.

(2) BX is twice as long as AX.

Thought this one was particularly brutal, then I read the OE and was kicking myself

This problem hinges on recognizing that the two triangles are similar because they share an angle and they are both perpendicular. In this case, the two triangles are perpendicular because the line formed by their respective radii bisects AB. So once we see that 90 degree angle and the shared angle, we can determine that they are similar.

Statement 1: We know that the radii are related as follows: piR^2 = 4*pi*S^2. So R/S = 2. Once we know the relationship of two sides and the length of the both bases of the triangles (10, the x-coordinate of the larger triangle), we can solve for the x-value of the point.

Statement 2: We know that BX is 2*AX. In this case, we know the relationship of two sides, so we can solve for the length of the base of the triangles and thus the x-coordinate of X.
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Re: In the figure above, the circles are centered at O(0, 0) and [#permalink]
Bunuel

we don't necessarily need statement 1 or 2 to infer that that XP =2OX right?

I kept flipping contemplating the internal section formula but then noticed that the two triangles appear to be similar- I saw that you applied
xp =2ox to statement 1?
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Re: In the figure above, the circles are centered at O(0, 0) and [#permalink]
Nunuboy1994 wrote:
Bunuel

we don't necessarily need statement 1 or 2 to infer that that XP =2OX right?

I kept flipping contemplating the internal section formula but then noticed that the two triangles appear to be similar- I saw that you applied
xp =2ox to statement 1?

We know that triangles AOX and BPX are similar from the stem but how can you get the ratio? Why not XP =3*OX or 2*XP =OX? Or any other ratio?
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Re: In the figure above, the circles are centered at O(0, 0) and [#permalink]
OE:

Before dealing with either statement, note that the line containing A and B is perpendicular to the radii OA and PB of the two circles, creating triangles OAX and PBX. Each triangle contains a right angle, and the angles at point X are also identical (since they are vertical angles). Therefore, triangles OAX and PBX are similar.

(1) SUFFICIENT: Let R stand for the radius of the larger (right-hand) circle, and r the radius of the smaller (left-hand) circle. Then, according to this statement, πR2 = 4πr2. Therefore, R2 = 4r2, and so R = 2r. In other words, PB = 2(OA), or, equivalently, PB/OA = 2.

Since the triangles are similar, the ratio of similar sides is equal: PB/OA = PX/OX. Therefore PX/OX also equals 2. The problem indicates that the length of OP is 10, so this is enough information to determine the lengths of PX and OX, which will allow you to determine the x-coordinate of X.

Don’t actually do this math, but here’s how: OP is 10 and the portion PX is twice as long as the portion OX. The length of PX, then, must be (2/3)(10) while the length of OX must be (1/3)(10). The x-coordinate of X, then, equals 10/3.

(2) SUFFICIENT: Since the triangles are similar, the ratio of similar sides is equal: BX/AX = PX/OX. According to this statement, BX/AX is 2, so PX/OX also equals 2. The problem indicates that the length of OP is 10, so this is enough information to determine the lengths of PX and OX, which will allow you to determine the x-coordinate of X.

Don’t actually do this math, but here’s how: OP is 10 and the portion PX is twice as long as the portion OX. The length of PX, then, must be (2/3)(10) while the length of OX must be (1/3)(10). The x-coordinate of X, then, equals 10/3.

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Re: In the figure above, the circles are centered at O(0, 0) and [#permalink]
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Re: In the figure above, the circles are centered at O(0, 0) and [#permalink]
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