In the xy-plane, point O is located at the origin, point A has coordin

Ans should be B.
The most important clue that we have is AO>AB
This restricts how far r can go.
AO from coordinates will be 5 units in length. So maximum value of AB < 5.
And all points are in first quadrant only so maximum value of r can be 8 (excluding 8).
So maximum area of tringle could be
1/2 * 8 * 3 = 12 (excluding 12)
Let me know if its not clear or if I am wrong somewhere.

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Saurabh I think you considered the height of the triangle in incorrectly


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By using Heron's Formula
Area of triangle is 1/2(x1(y2-y3)+x2(y3-y1)+x3(y1-y2))
Consider statement 1 which is not enough
Now consider statement 2 which still does not gives us all three coordinates of vertices.
Thus, C is the answer.


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comprehensive soln anyone folks?

To add to my solution above,
Algebraically we can show how r should be less than 8.
We know that length of a line between two coordinates is equal to sqrt of ( (x1-x2)^2 + (y1-y2)^2 )
So from AO > AB we get:
AO^2 > AB^2 (since both sides are positive )
So (4-0)^2 + (3-0)^2 > (r-4)^2 + (3-0)^2
Simplifying :
(r-4)^2 < 16
This gives us 0 < r < 8

Let me know if its not clear or if I am wrong somewhere.



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oh wow...took 3 minutes just to make sure i don't need 1...
a truly 700 level question

Hey! Why is it excluding 8 here? maximum value of r can be 8 rite? area can also be 12, which is the maximum. Pls throw some light here!

Thanks,
Uma

While I understood the algebraic approach to deduce that r should be <8. The length of the 3rd side should be: greater than 1 and less than 9. Is there any other way to deduce that r<8?

Bunuel: Please throw some light here.

\({S_{\Delta ABO}} = \frac{{r \cdot q}}{2}\,\,\mathop < \limits^? \,\,12\,\,\,\,\,\, \Leftrightarrow \,\,\,\boxed{\,\,r \cdot q\,\,\mathop < \limits^? \,\,24\,\,}\)




\(\left( 1 \right)\,\,r = 7\,\,\,\left\{ \matrix{\\
\,{\rm{Take}}\,\,q = 1\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr \\
\,{\rm{Take}}\,\,q = 4\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr} \right.\)


\(\left( 2 \right)\,\,\left( {p,q} \right) = \left( {4,3} \right)\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,r \cdot q < 8 \cdot 3\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle\)

\(\left( * \right)\,\,4 = p > {r \over 2}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,r < 8\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

Hi, TheMechanic!

I will "intrude" to help. I hope Bunuel and yourself accept my apologies, if needed.

The perpendicular bisector of line segment OB is the set of all points (in the plane) that are equidistant of O (the origin) and B (given).

Hence AO > AB occurs if, and only if, A is "nearer" B, i.e., "to the right" of the perpendicular bisector mentioned, as shown in my figure (post above).

That´s why we must have p greater than r/2 , therefore (in sttm 2)), r/2 less than 4, i.e., r less than 8.

I hope you got things clearer now.

Regards,
Fabio.

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