If PQ is a diameter of the circle above, and the lengths of AX, BX, PX

im not seeing how the two triangles are similar. i get that AXP and BXQ are vertical angles, but cant find any relation between the other two angles.

The length of AB is at 8, 10 or 17.

the condition QX > AB is only possible if QX = 16
Why is this so?
Cant QX also be any integer greater than 16?

Not understanding the steps here to get to 17 as the diameter........

how can we say that the two triangles formed are similar?? we are given that only measure of one angle is same in both triangles which is insufficient to prove congruency.. please explain

Hi there, you seem to be not the only one with confusion on this so I will do my best to explain.

Firstly it primarily has to do with the fact that vertices of each triangle is formed at the ends of the diameter of the circle they are inscribed within. Since the line AB intersects the circle (and its diameter) it is sufficient to say that these angles are similar within the triangles as well. If two angles of triangles are the same measurement then the third must also be the same making them similar triangles.

A more visualization approach that may help illuminate this fact is as follows. If you were to draw a line exactly perpendicular to the diameter at exactly point X then fold the circle at both the diameter and the perpendicular line you would have two triangles perfectly inscribed with the longest sides (BQ and AP) exactly parallel with each other. All of the angles would either overlap or be the same giving us similar triangles for comparison. You just have to remember that similar triangles can be manipulated visually to fit on-top of one another.

This is a very sorry if this does not clarify it. This is a very tricky problem and I honestly doubt that this is a real question that you could expect to see on the actual test.

I will give it a shot. In general a secent(segment ) rule says that the product of cords that intersect are equal. Proof does stem from the fact that the triangles in discussion are similar. Google it and you will find the proof.

Here is one site you can explore different circle rules. https://www.regentsprep.org/regents/math ... gments.htm

In nutshull we have a relation of intersection cords that will be helpful to remember on GMAt rather than proof.
Same has been used here : Px. QX = AX.BX. Based on this theorem the question is solved.

I think it's pretty well explained from the Data Sufficiency angle. So, not solving the entire problem. But just adding a bit on the first part of the explanation related to the concepts of Intersecting Chords & Similarity. To understand the problem in a better way let us remember the following rules/ theorems.

1) Theorem related to Intersection of Chords: When two chords intersect each other inside a circle, then each is divided into 2 segments and the products of their segments are equal.
Thus in this question, when we consider the chords AB and PQ (note: Diameter is the largest chord of a circle), they intersect at X. Hence AX . BX = PX . QX
You can straightaway apply this to proceed with this question.

2) Theorem related to Angles subtended by an Arc on the Circle: Angles subtended on the circumference in the same segment of a circle by the same arc are always equal (note: These are call Internal Angles, and are half of the Central Angle subtended by the same arc at the centre of the circle).
Thus in this question if we consider the arc PA, it subtends two angles - one at B and another at Q.
Hence, angle(PAB) = angle(PQB)

Now if you consider the two triangles, PAX and QBX, then they have
angle(PAB) = angle(PQB) ..... as shown above
angle(PXA) = angle(QXB) ..... vertically opposite angles
And hence they are SIMILAR (note: two triangles are similar if all the three angles are equal and in fact if two sets of angles are equal, then automatically the third set of angles must also be equal).
Thus their "Corresponding Sides" will be in the same Ratio (note: by Corresponding sides I refer to the sides opposite to the equal angles in two similar triangles).
So, AX/ QX = PX/ BX
i.e. AX . BX = PX . QX
Again we arrive at the same results as in Point 1 and can proceed from there.

Although I would prefer a student to straightaway go for method 1, I thought of discussing the method 2 since the concept of Similarity was already discussed in context to this problem and I think all can develop good insights by understanding this approach.

hi there,

Can you please help me understand why can't the ax.bx be equal to 1x16 (or vice versa) why does it has to be 8x2 (or vice versa)?

Thanks in advance!

Sure "neeraj609".
Now we know, AX . BX = PX . QX
From St 1, we know AX . BX = 16; So, PX . QX is also 16.
Since all these have integer values, hence options can be (1, 16), (2, 8) or (4, 4)
Our interest is to find the Diameter (PQ = PX + QX)
With all 3 possible values of (PX, QX), PQ can also have 3 values: 17, 10 and 8
Thus, Statement 1 is not sufficient.

St 2 doesn't give us anything at all. So even that is not sufficient by itself.

Note, AB can also have 3 values: 17, 10 and 8 (with varying integer pairs following AX.BX = 16)
And possible values of QX are 1, 2, 4, 8, 16 (note: corresponding values of PX will be 16, 8, 4, 1 to ensure PX.QX = 16)

If we combine St 1 and St 2, to satisfy QX > AB, QX must be 16 (and AB must be 8 in that case).
When QX = 16, PX will be 1.
So diameter PQ will be 17.
And we can find the Area of the Circle.

Hope it helps. If not please write back.

# This is a brilliant problem that works on so many concepts of Geometry all at the same time and is very much within the scope of GMAT. But it's actually difficult to solve in real-time exam.

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