# 12.4 Problems. Excerpt from "Introduction to Geometry" 2014 AoPS Inc. Copyrighted Material CHAPTER 12. CIRCLES AND ANGLES

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1 HTER 1. IRLES N NGLES Excerpt from "Introduction to Geometry" 014 os Inc. onider the circle with diameter O. all thi circle. Why mut hit O in at leat two di erent point? (b) Why i it impoible for to hit O in three di erent point? Hint: 530 (c) Let the point where hit O be and. rove that \O = \O = 90. (d) rove that and are tangent to O. (e)? Now for the tricky part proving that thee are the only two tangent. Suppoe i on O uch that i tangent to O. Why mut be on? Hint: 478 (f) Why doe the previou part tell u that! and! are the only line through tangent to O? 1.3.6? O i tangent to all four ide of rhombu, = 4, and = 15. rove that and meet at O (i.e., prove that the interection of the diagonal of i the center of the circle.) Hint: 439 (b) What i the area of O? Hint: roblem roblem roblem 1.19: quadrilateral i aid to be a cyclic quadrilateral if a circle can be drawn that pae through all four of it vertice. rove that if i a cyclic quadrilateral, then \ + \ = 180. roblem 1.0: In the figure, i tangent to the circle and biect \E. Furthermore, = 70, E = 50, and \QE = 40. In thi problem we determine the meaure of the arc from to E that doe not include point. Find. 70 Q 40 E 50 (b) Find \. (c) Find. (d) Finih the problem. roblem 1.1: Three congruent circle with radiu 1 are drawn inide equilateral 4 uch that each circle i tangent to the other two and to two ide of the triangle. Find the length of a ide of 4. Hint: 48! roblem 1.: i tangent to both circle O and circle. Given that O = 40, and the radii of circle O and are 31 and 7, repectively, find. Hint: 548, 67 O 3 opyrighted Material

3 HTER 1. IRLES N NGLES Excerpt from "Introduction to Geometry" 014 os Inc. Now that we have all the other arc of the circle, we can find our deired arc by ubtracting from 360 : = 00. oncept: Label all the given information in the problem, and all the information you find a you find it. Thi will help you dicover new fact about the problem. For example, your final diagram in the lat problem might look like the figure below. Notice that our variou arc meaure and angle are marked, and that the equal angle at are marked equal Q E roblem 1.1: Three congruent circle with radiu 1 are drawn inide equilateral 4 uch that each circle i tangent to the other two and to two ide of the triangle. Find the length of a ide of 4. Solution for roblem 1.1: We need to create imple figure to work with, o we tart by connecting the center of our circle and drawing radii to tangent point. (Note that connecting the center of the circle i the ame a drawing radii to where the circle are tangent.) Since W i a rectangle (becaue W =, W k, and \W = 90 ), we have W = =. Hence, we need only find W to finih, ince i the ame a W. We draw to build a right triangle and note that thi egment biect \ becaue circle i tangent to both and (and hence W it center i equiditant from them). Since \W = (\)/ = 30, 4W i a triangle. Thu, W = W p 3 = p 3. Finally, we have = +W +W = + p 3. oncept: When you have tangent in a problem, it often very helpful to draw radii to point of tangency to build right triangle. When you have tangent circle, connect the center. (In fact, if you have multiple circle in a problem, connecting the center will ometime help even when the circle aren t tangent.)! roblem 1.: i on circle O and on circle uch that i tangent to both circle. Given that O = 40, and the radii of circle O and are 31 and 7, repectively, find. 34 opyrighted Material

4 Excerpt from "Introduction to Geometry" 014 os Inc ROLEMS Solution for roblem 1.: We tart with the uual egment to draw:, the radii to point of tangency, and the egment connecting the center. We till don t have a right triangle to work with, but we do know that radii O and are both perpendicular to tangent a hown. Since both O and are perpendicular to the ame line, they are parallel. We make a right triangle and O a rectangle by drawing a line through parallel to. = = 7, o O = O = 4. Since O = 40, we have = 3 from right triangle 4O. Since i a rectangle, we have = = 3, o the length of the common tangent i 3. i called a common external tangent of the two circle. an Exercie, you ll find the length of the common internal tangent, too. roblem 1.3: Median M of 4 ha length 8. Given that = 16 and = 9, find the area of 4. Solution for roblem 1.3: When we draw the figure and label all our length, we ee that M = M = M = 8. Therefore, a circle centered at M with radiu 8 goe through all three vertice of 4. Since i a diameter of thi circle, \ i incribed in a emicircle and therefore mut be a right angle. So, = p = 5 p 7. 4 i a right triangle, o it area i half the product of it leg: [] = ()() = (9)(5 p 7) = 45 p 7. M Uing thi ame reaoning, we can alo prove thi important fact: Important: If the length of a median of a triangle i half the length of the ide to which it i drawn, the triangle mut be a right triangle. Moreover, the ide to which thi median i drawn i the hypotenue of the right triangle. We can alo look to thi problem for ome important problem olving technique: oncept: When tuck on a problem, alway ak yourelf Where have I een omething like thi before? In roblem 1.3, we have a median that i half the ide to which it i drawn. Thi hould make u think of right triangle, ince the median to the hypotenue of a right triangle i half the hypotenue. Then we go looking for right triangle. oncept: lway be on the lookout for right triangle. roblem 1.4: ide of 4. Let be the point where the incircle of 4 meet. Find in term of the 35 opyrighted Material

5 HTER 1. IRLES N NGLES Excerpt from "Introduction to Geometry" 014 os Inc. Solution for roblem 1.4: We tart by labeling the equal tangent from the vertice a hown in the diagram. We want to relate x to the ide of the triangle, o we write the ide of the triangle in term of x, y, and z. We let the ide of the triangle be = c, = b, and = a and we have: y x x z = c = x + y = b = x + z = a = y + z y z We want x in term of a, b, and c. dding the three equation will give u x + y + z, which we can ue with y + z = a to find x: a + b + c = (x + y + z) o x + y + z = a + b + c. We can then ubtract the equation y + z = a from x + y + z = (a + b + c)/ to find x = a + b + c a = a, where i the emiperimeter (half the perimeter) of the triangle. Similarly, y = b and z = c. oncept: Symmetric ytem of equation can often be eaily olved by either multiplying all the equation or adding them. Important: The length from the vertice of 4 to the point of tangency of it incircle are given a follow: = = a = = b = = c b a a c where = c, = b, and = a, and the emiperimeter of 4 i. b c Exercie roblem 1.4.3, 1.4.4, 1.4.5, and are very important relationhip that you ll be eeing again in your tudy of more advanced geometry. e ure to pay pecial attention to them I every quadrilateral cyclic? 1.4. rove the following about cyclic quadrilateral: (b) (c) cyclic parallelogram mut be a rectangle. cyclic rhombu mut be a quare. cyclic trapezoid mut be iocele. 36 opyrighted Material

Section 3.4 Pre-Activity Preparation Quadrilateral Intereting geometric hape and pattern are all around u when we tart looking for them. Examine a row of fencing or the tiling deign at the wimming pool.

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