<p>View</a> image: Pand Q</p>
<p>T is a point (not shown) on semicircular arc SRP. T is different from S and P.</p>
<p>Which is bigger: (A)The distance between point Q and T OR (B)3?</p>
<p>This is from a 1995 SAT exam, btw. </p>
<p>I know that A is bigger, obviously, but I can't seem to get a well-founded explanation for why it's correct.
So, what values should QT be between?</p>
<p>I remember this problem from way back. I can give you a math reason why QT is greater than 3. But to find the largest possible value of QT looks like it might be a longer problem. For now:</p>
<p>First find the midpoint of the hypotenuse, which is also the center of the circle. Call it C.</p>
<p>Then, put a point on the semicircle that is co-linear with C and Q. Call that D. We are going to consider possible locations for T on the semicircle between S and D. Look at the triangle QST and in particular the angle at S. First you need to convince yourself that if point T is all the way at point D, then that angle at S is inscribed in a semicircle so it is 90 degrees. And if T is between S and D, then the angle is greater than 90.<br>
Now, if QT were less than or equal to QS, then angle S would be less than or equal to the angle formed at T. But that gives you two angles in the triangle that are at least 90 degrees. </p>
<p>But what about when T is further along the arc, past D? Then, the same argument applies, starting with the other side of the figure. So in that case, QT has to be greater than 4!</p>
<p>I hope this makes some sense. It’s obviously more than they expected you to do back then – you got full credit just for seeing that it had to be so. I remember that I needed to play with Geometer’s Sketch Pad to realize why it had to be true. </p>
<p>BTW, I am impressed at the thoroughness of your prep to have re-uncovered this old gem.</p>
<p>Thanks for the reply.</p>
<p>Btw, just to make this clear. When you say:
“Look at the triangle QST and in particular the angle at S. First you need to convince yourself that if point T is all the way at point D, then that angle at S is inscribed in a semicircle so it is 90 degrees.”
Angle S is not inscribed in semicircle SRP, which is the only obvious semicircle I can see. So, you’re talking about a semicircle with diameter QCD, right? If so, then for me that makes sense why QD and QT will always be bigger than QS or QP.</p>
<p>^Right. Not the drawn semicircle, but the one you have described.</p>
<p>Doing this really quick from a phone … my take is that all that is needed to rotate QS visually. The information that Q is a right angle is also telling that the highest value for QS is 3. When rotating, all values will be smaller than three as the angle Q is reduced from 90 degrees, and will never cross the semi-circle. </p>
<p>In a way, this problem is somehow similar to a more recent problem of finding the largest possible area of a right triangle when two sides are given. Knowing that a right triangle always yields the highest area was the ticket to answer that latter one.</p>
<p>^Not quite seeing what you mean. Maybe upload a diagram later? (Or not – it’s a really dated problem and one that I bet they wish they could take back. It doesn’t seem to be testing anything of interest.)</p>
<p>Another way to think of it:</p>
<p>Draw the circle of radius 3 centered at Q. Notice that all points on arc SRP (except S) lie outside the circle of radius 3. Therefore any point T you pick on arc SRP will satisfy QT > 3.</p>
<p>^That’s certainly true. But if your intuition failed you, how would you know to draw the new circle INSIDE the one they gave you? To draw it properly, you already have to know the answer to the question!</p>
<p>I know, I was just suggesting a quick proof of why QT > 3. On some harder problems it helps to draw random lines and circles.</p>
<p>Again, there is some information one can rely on to decide to construct that additional circle. The fact that the triangle is a right one yields the information of the side being 3, and that this is its maximum value when rotating. That is what I meant by “rotating” the short side of the right triangle. At no point, will the rotation reach the given semicircular arc. This is the same approach as drawing a full circle from the right angle point (Q.)</p>
<p>By the way, a nice current SAT question could be … what is the area under the semicircular arc. Or, even better, what is the ratio between the half-circle and the triangle in terms of surface. :)</p>
<p>This is a nice problem. Here is a visual proof of why QT>3 and the maximum value of QT is equal to 5. Let me know if there is anything I missed in the logical flow. </p>
<p>[Circle:</a> Max Value of QT - Dabral’s library](<a href=“http://www.screencast.com/t/PtdLZtmw4xar]Circle:”>http://www.screencast.com/t/PtdLZtmw4xar)</p>