<p>1) A polygon is inscribed in a circle. If some diameter of the circle divides the polygon into two regions of equal area, which of the following must be true ?</p>
<p>I) The polygon is regular
II) The center of the circle is inside the polygon
III) The polygon is symmetrical about some diameter</p>
<p>A) None
B) I only
C) II only
D) III only
E) II and III only</p>
<p>2) An initial deposit of $2,000 is placed in an account that pays 5.5 percent annual interest, compounded quarterly. If no other deposit or withdrawals are made, after how many quarterly interest payments does the amount in the account first exceed $4,000 ?</p>
<p>Assume that the circle is in the x-y plane with its center at the origin. Think of points on the circumference of the circle as making an angle of 0 degrees to 360 degrees with the x axis. Measure counter clockwise.</p>
<p>Hopefully this will help you visualize the following:</p>
<p>Consider a triangle ( e.g. a polygon) with vertices on the circumference of the circle at 120, 180, and 190 degrees. By definition of inscribed this triangle is inscribed in the circle.</p>
<p>Note
(1) The triangle is not regular (i.e. it is not equilateral). </p>
<p>(2) The center of the circle is not inside the triangle.</p>
<p>(3) The triangle is not inherently symmetric about any line.</p>
<p>Yet you should be able to see that some diameter that intersects the circumference at a point between 120 and 190 degrees will divide the triangle into two equal areas. The easiest way to convince yourself of that is to observe that the diameter at 120 degrees does not divide the triangle at all. Nor does a diameter at 190 degrees. One of the diameters in between must.</p>
<p>1) I is false. Take a rectangle that is bisected by the diameter. II is false. Given the diameter and the center as the midpoint of the diameter, construct a quadrilateral symmetrical about the diameter that is on one side of the diameter and thus does not contain the center. I think III is also false. Construct a small square with one side on the diameter. Now continue one of the sides of the square past the diameter and let this form an altitude of a right triangle with base on the diameter (shared with the square) such that the area of the triangle and square are equal. The resulting quadrilateral should have no axis of symmetry.</p>
<p>2) This is just plugging into a formula: A=P(1+r/n)^(nt) P=2000, r=0.055, n=4</p>
<p>4000=2000(1+0.055/4)^(4t). We are solving for 4t which is the number of times the interest is compounded. 2=(1.01375)^(4t), ln2=4t ln1.01375, 4t=50.7. Round to 51 because that is when we will pass the 4000.</p>
<ol>
<li><p>C
I - false, think of isosceles traingle
II - false, draw a polygon outside of the center of the circle, like a trapezoid
III - true, first 2 are def wrong. this one must be right. also, i can’t think of any instance where it wouldn’t. not sure though.</p></li>
<li><p>B On a calculator 2000(1.055^x). 52 is the first number where it is above 2000.</p></li>
</ol>
<p>1.
I and II are false (consider an isosceles triangle with vertex angle >90 deg)</p>
<p>III is tricky. The SAT never requires you to prove anything (if III is true, you have to show that it holds for every such polygon, or just blindly assume it is true). It is usually easier to start looking for counterexamples.</p>
<p>Consider right triangle ABC with AB = 3, BC = 4, AC = 5. If BD is a diameter on the circle, then BD intersects the midpoint of AC and therefore divides the triangle into two smaller triangles of equal area. However, triangle ABC does not have any lines of symmetry. Hence III is false, and the answer is A.</p>
<p>I think III is false, I can actually draw a polygon that is not symmetrical, yet equal when cut in half. If you draw a really weird 12+ sided polygon, you can draw it in a way so that it can divide in half equally. With this you can have a crazy shape that can’t be symmetrical in any case. </p>
<p>I got A for the 1st question and B for the 2nd question.</p>
<p>On a side note, you can rigorously prove that there must exist a diameter that divides the polygon into two equal areas. Suppose that, on the contrary, that this isn’t true. Then any diagonal divides the polygon into two areas X and Y, such that X > A/2 and Y < A/2 (A denoting the area of the whole polygon).</p>
<p>Suppose we fix the semicircles where X and Y are. If we rotate the diagonal by 180 degrees, it turns out that X becomes Y and Y becomes X. The areas of X and Y are continuous functions of the angle (this can be tricky to prove), and by the intermediate value theorem, there must exist a diagonal such that X = Y = A/2. Therefore any cyclic polygon has a diagonal that bisects the area.</p>
<p>Clearly, not all polygons have line symmetry, so III is false.</p>