<p><a href="http://i%5B/url%5D">http://i</a>. imgur .com/mQ7woTb.png (remove spaces)</p>
<p>Answers are A and C</p>
<p>the first one:</p>
<p>According to the problem, you can draw a figure with R as the midpoint of Q and S, and P directly above R. From that, you can draw triangles connecting the points. In a right triangle, the hypotenuse must be the longest side, so it follows that PQ (the hypotenuse) would be greater than PR (a leg of the triangle)</p>
<p>the second one:</p>
<p>okay, this one’s a bit hard to explain without me being right next to you, but I’ll try.</p>
<p>According to the problem, segment AB must be an integer. That means, the x - value of y = x^2 must be a whole number. If it wasn’t, then AB wouldn’t be an integer. Whole number x-values for y = x^2 could be (1,1) (2,4) (3,9) (4,16) (5, 25) … etc.</p>
<p>From this it follows that you must find what k can equal at those points, so you plug each one of those points into y = k - x^2 . For example, for (1,1), you do 1 = k - 1. From there, you get k = 2. Therefore, k = 2 IS a possible value. After plugging in the rest of the points, you get that k can equal 2,8,18, 32, 50… All of those are in the answers except for C. </p>
<p>Sorry if this is kinda hard to understand, I tried my best to explain it easily.</p>
<p>For problem 2, the x-coordinate of point B (the intersection of the parabolas) is the positive real number x such that x^2 = k - x^2, so 2x^2 = k.</p>
<p>x is an integer (given), so k must be 2 times a perfect square. All of the answer choices are of this form except C.</p>
<p>Thanks for the replies.</p>
<p>How do you see problems in these ways? These are relatively simple concepts and I know them, but there was no way I could have known to make two triangles out of P Q R and S, or see that k - x^2 = x^2. Does it just come with practice?</p>
<p>It’s mostly problem-solving strategy and skill. Use all the information you know to write equations or claims that must be true. For many geometry problems, drawing important lines or points can help.</p>