<p>Can someone explain these math problems?</p>
<p>1) A circle and a triangle are drawn on a piece of paper. Which of the following is the set of the number of possible points that are common to both the circle and the triangle?</p>
<p>(A) {0,2}
(B) {0,1,2,3}
(C) {2,4,6}
(D) {1,2,3,4,5,6}
(E) {0,1,2,3,4,5,6}</p>
<p>2) Two parallel lines are drawn in a plane. Two additional lines are drawn in the same plane, in such a way that there are exactly three diffrent intersection points between the four lines. Neither of the additional lines is parallel to the two original lines. Into how many non-overlapping regions do these lines divide the plane?</p>
<p>(A) Five
(B) Six
(C) Seven
(D) Eight
(E) Nine</p>
<p>The answer is (E) for both of them.</p>
<p>These are from a Kaplan practice test so can someone also tell me if they have seen these kinds of problems on the real test?</p>
<p>I have never seen anything like this on the SAT. These are much closer to being brain-teasers than the SAT ever gets.</p>
<p>The first one
0: obviously, it’s possible that they don’t touch at all.
1: the edge of the circle touches a corner
2: the circle is pushed slightly towards the triangle from (1) so it crosses twice near the corner
3: the circle is inscribed in the triangle
4: slightly adjust the triangle from (3) by shortening one side so the circle overlaps one side, but still just touches the other two
5: repeat changes in (4) for a second side
6: repeat changes in (4) for the third side, so the circle overlaps all three sides</p>
<p>I must be missing something as well for the second one, because all I can think right now is that any non-parallel lines in a plane MUST cross, so I don’t see how there can only be three intersection points. Maybe I’m just being an idiot, sorry.</p>
<ol>
<li>It’s possible to draw a circle intersecting a triangle at six different points (try it). It’s also possible to draw a circle that does not intersect the triangle, so we know that {0,6} is in the set. The only possible answer is E.</li>
</ol>
<p>Note: To obtain the other values, draw the circle so that it is tangent to one or more of the sides of the triangle.</p>
<ol>
<li>Draw it out. The third line must create two intersections. Then the fourth line must create exactly one more intersection, therefore it must intersect at one of the other two intersections (creating a triangle). Count the number of regions, you should obtain nine.</li>
</ol>
<p>Ok thanks guys, I get the first one but the second one is still a little vague. Has anyone seen these kind of problems on the real SAT?</p>
<ol>
<li> Draw an x.</li>
<li> Draw a line through the center of the x so that you have three lines intersecting at one point.</li>
<li> Draw a line parallel to the line you drew in step 2.</li>
</ol>
<p>This will give you the figure described in the problem. The only way to have only three points of intersection with four lines is if there is a point at which three of the lines intersect.</p>
<p>And yes, I have seen CB questions like these.</p>
<p>For #2, just draw it exactly the way neatoburrito or I described it.</p>