<p>Question:</p>
<p>In the figure above, each of the four large circles is tangent to two of the other large circles, the small circle, and two sides of the square. If the radius of each of the large circles is 1, what is the radius of the small circle?</p>
<p>(A) 1/4
(B) 1/2
(C) (Square root(2)-1)/2 (approximately 0.207)
(D) Square root(2) -1 (approximately 0.414)
(E) Square root(2)/2 (approximately 0.707)</p>
<p>Image:
<a href="http://img146.imageshack.us/img146/6076/circlemathquestionpu7.jpg%5B/url%5D">http://img146.imageshack.us/img146/6076/circlemathquestionpu7.jpg</a></p>
<p>Question:</p>
<p>If s is the sum of 3 consecutive odd integers and n is one of the 3 odd integers, which of the following could be true? </p>
<p>(A) s=3n
(B) s=3n+3
(C) s=3n-3
(D) s=6n
(E) s=6n+1</p>
<p>Thanks</p>
<p>1st one is really hard I want to say its (E)
2nd one is definitely (A)</p>
<p>For the first one, draw the right triangle which has the following 3 vertices:</p>
<p>(1) the center of the small circle
(2) the center of one of the larger circles (let’s say the one at the bottom right just to be specific)
(3) below vertex (1) and to the left of vertex (2) so that a right triangle is formed.</p>
<p>The legs of these triangles each have length 1 (they are the same length as the radius of the big circle). Thus the hypotenuse has length sqrt(2). </p>
<p>Note that the hypotenuse of this triangle is the sum of the radius of the big circle and the small circle. Thus the small circle has radius sqrt(2) - 1, choice (D)</p>
<p>Quick Tip: In “hard” geometry problems with circles look for “key points” on the circle to draw radii.</p>
<p>This question has a bunch of “key points”. They are all the intersection points (such as the intersection of any 2 circles in the picture). Note that I would consider the points of intersection of the circle with the square key points, but they are not used in solving this problem.</p>
<p>We can do the second one by picking numbers. Let’s let n=5. Then there are 3 possibilities for the consecutive odd integers:</p>
<p>1st possibility: 1,3,5 Sum=9
2nd possibility: 3,5,7 Sum=15
3rd possibility: 5,7,9 Sum=21</p>
<p>Now we start plugging n=5 into the answer choices until we get an answer of 9,15, or 21</p>
<p>(A) s=3n=3*5=15</p>
<p>This works, so the answer is choice (A).</p>
<p>Algebraic solution: Note first that I recommend the above method for solving this problem. I’m just including this for completeness.</p>
<p>There are 3 possibilities for the consecutive odd integers:</p>
<p>1st possibility: n-4, n-2, n Sum=3n-6
2nd possibility: n-2, n, n+2 Sum=3n
3rd possibility: n, n+2, n+4 Sum=3n+6</p>
<p>Now just look at the answer choices. Only choice (A) matches.</p>
