Randolph High School Math League Page 1. Similar Figures

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1 Randolph High School Math League Page 1 1 Introduction Similar Figures Similar triangles are fundamentals of geometry. Without them, it would be very difficult in many configurations to convert angle information to length information (e.g. through similarity) and vice versa. t the same time, though, similar triangles are often the most difficult things to spot without prior geometric intuition. In this workout, the goal is to strengthen this intuition. 2 Warm-up Problems The one thing that is suboptimal about teaching a geometry topic to a math club is that, unlike algebra, geometry is not reiterated in the high school mathematics curriculum. s a result, there may be various facets of geometric topics that may have fizzed away over time. To refresh your memories, try the following problems. The first two are more basic in nature, while the third is a tad bit harder (but still quite doable). If there are any gaps in knowledge present from geometric atrophy, let me know and I ll be glad to help. 1. [HSM 1995] In, = 90, = 6 and = 8. Points and are on and, respectively, and = 90. If = 4, then = () 5 () 16 () 20 () 15 2 () 8 2. How many equilateral triangles of side length 1 can fit inside an equilateral triangle of side length? () 6 () 7 () 8 () 9 () 10. has = 12, = 1, and = 15. Points X and Y are placed on and respectively such that XY =. If XY = 6, what is X + Y? () 4 () 6 () 8 () 10 () 12 Utilizing Similar Triangles In all of the above problems, the similar triangles were fairly easy to spot: an angle condition was given (e.g. parallel or antiparallel lines), this condition was used to find similar triangles, and from there lengths of segments in the diagram could be found. In competitions (or in real life for that matter) the situations are not as simple. In this regard, we will dive straight into a few problems; the hope is that you will pick up on a technique or two throughout the way. xample 1 (M ). Points,,,,, and F lie, in that order, on F, dividing it into five segments, each of length 1. Point G is not on line F. Point H lies on G, and point J lies on GF. The line segments H, J, and G are parallel. Find H/J. G H J F Solution. t first glance, this diagram seems quite complicated - the key is to pick it apart. More specifically, we have three sets of parallel lines here: H, J, and G. From H G we see H G, so

2 Randolph High School Math League Page 2 HG G = = 1 J. In a similar fashion, JF GF, so G = F F = 1 5. Now we see that this problem is not so bad after all! Indeed, manipulating the ratios so that the G terms cancel gives H J = ( H G ) ( ) G = 1 J 5 = 5. xample 2. On square, points and F are constructed on and respectively such that = and F = 2F. Segment F intersects and at P and Q respectively. If =, what is P Q? Solution. Note that from the condition on point F we have F = 2, so F = = 1. It now suffices to use similarity to determine P Q. First examine lines and F. They form two triangles, F Q and Q, which are similar to each other. Their ratio of similitude is F = 2, so Q = x and F Q = 2x for some x. Now remark F = x+2x = 5x = 1, so x = 1 5 and Q = 5 1. In a similar manner, we examine and F. Note that F P P, so P P F = F = /2 2 = 4. Now let P = y and P F = 4y; then F = 7y, which implies y = 1 7. Thus, P = 7 1. Putting everything together, we get P Q = Q P = = onstructing uxiliary Lines Often times, problems will be made easier by constructing additional lines within the diagram. If the right lines are drawn, similar triangles can be exploited. Here is a short example illustrating this idea: xample (Math League HS ). The area of square is 1. s illustrated at the right, diagonal is extended its own length to point. How long is? Solution. Let M be the midpoint of. Then since = 1 and M = 1 2, Pythagorean Theorem gives M = Now remark that M, so M = M = 2 = = 2 M = 5. This is not the only simple solution to the problem. We could have also dropped a perpendicular from to and scaled upward before utilizing Pythagorean Theorem - that works just as well. In general, if you feel an auxiliary line is necessary, there are three types of lines that work better than others: Lines parallel to others in the diagram. s we saw in the previous problem, this creates similar triangles that may be useful. Perpendiculars. Right triangles are very powerful tools since they come equipped with their own set of theorems. rop altitudes when you need information regarding perpendicular distances or when you need to set up similar right triangles. (n altitude from the vertex angle of an isosceles triangle can be especially useful, since its foot also doubles as the midpoint of the base!) Radii of a circle. The most recurring case of this occurs with points of tangency: if l is a line tangent to a circle with center O at point, then O l. (Remember this fact!)

3 Randolph High School Math League Page 5 Problem from the IM efore we move on to our final example, here is a seemingly-silly problem that demonstrates an important concept with similar triangles: xample 4. Two triangles, I and II, are known to be similar. Furthermore, it is known that the length of the median to the longest side of I is 6 while the length of the median to the longest side of II is 9. What is the ratio of the area of I to the area of II? Solution. The ratio of similitude between I and II is 6 9 = 2. Therefore the ratio of their areas is ( 2 )2 = 4 9. Notice that despite the fact I was given nothing about the actual side lengths of either triangle, I was still able to determine the ratio of similitude between them. This is because when a triangle is scaled upward or downward, all components of the triangle (not just the sides) are scaled. In particular, I used lengths of medians as corresponding segments between two triangles. ven the fact that the segments are medians does not change the problem; for all I know, I could have been given the distance between the symmedian point 1 of the intouch triangle 2 and the reflection of the Fermat point of the orthic triangle 4 about the nine-point center 5 for both triangles and would have been able to make the same conclusion. We ll end this lecture with a more challenging problem from the 2015 merican Invitational Mathematics xam, which took place exactly one week ago. The question below was placed at the #7 spot, meaning that it is a medium-level problem when compared to the others that appeared on the test. xample 5 (IM 2015). In the diagram below, is a square. Point is the midpoint of. Points F and G lie on, and H and J lie on and, respectively, so that F GHJ is a square. Points K and L lie on GH, and M and N lie on and, respectively, so that KLMN is a square. The area of KLMN is 99. Find the area of F GHJ. N M G H K L F J Solution. Let X = HG and Y =. Note that HG, so HX = Y. ombining this with XH = Y = 90 gives Y HX. Now note that because the two squares F GHJ and KLMN are inscribed inside these triangles, their ratio of similitude is the same as the ratio the similitude of the aforementioned [F GHJ] two triangles. In other words, [KLMN] = ( ) 2 H. (Here [X] denotes the area of figure X.) Now let s be the side length of square and s 1 the side length of square F GHJ. Note that since F J GY H Y, we have F = 1 2 s 1 and GY = 2s 1. Thus 1 2 s 1 + s 1 + 2s 1 = 7 2 s 1 = Y = s 5 = s 1 = s. This implies H = ( 2 5 )s 1 = 4 7 s. Finally, we trivially have H = 7s, so [F GHJ] = [KLMN] ( ) 2 ( ) 2 s = 99 = H s/7 9 = the point where the reflections of the medians across the respective angle bisectors concur 2 the triangle whose vertices are the tangency points of the incircle with the three sides of the original triangle the point F inside for which F + F + F is minimized 4 the triangle whose vertices are the feet of the three altitudes of the original triangle 5 the center of the circle passing through the midpoints of the three sides, among other points; also the midpoint of the segment connecting the circumcenter and orthocenter of the triangle

4 Randolph High School Math League Page 4 6 Problems 6.1 Problem Set 1. Let points,,,,, F lie in the plane such that F. Let H a denote the unique point on for which H a, and define H d similarly. If H a = 5, = 6, and H d = 10, what is F? 2. [HSM 1990] Let be a parallelogram with = 120, = 16 and = 10. xtend through to so that = 4. If intersects at F, then F is closest to 4 F () 1 () 2 () () 4 () 5. [dapted from HMMT 2007] We are given four similar triangles whose areas are 1 2, 2, 5 2, and 7 2. If the smallest triangle has a perimeter of 4, what is the sum of all the triangles perimeters? 4. [M ] xternally tangent circles with centers at points and have radii of lengths 5 and, respectively. line externally tangent to both circles intersects ray at point. What is? 5. [M ] Rectangle has = 4 and =. Segment F is constructed through so that F, and and lie on and F, respectively. What is F? 6. [Wikipedia, et. al.] closed planar shape is said to be equiable if the numerical values of its perimeter and area are the same. For example, a square with side length 4 is equiable since its perimeter and area are both 16. Show that any closed shape in the plane can be stretched or shrunk to become equiable. 7. [HSM 1986] In, = 8, = 7, = 6 and side is extended, as shown in the figure, to a point P so that P is similar to P. The length of P is P () 7 () 8 () 9 () 10 () Let be a triangle with = 1, = 14, and = 15. Square Y X is erected outside. Segment X intersects at point P, while Y intersects it at point Q. etermine the length of P Q. 6.2 Problem Set 9. [Mandelbrot ] Suppose that is a trapezoid in which. Given, bisects angle, and area() = 42, then compute area(). 10. [ITMO 2012] On the sides of a triangle right angled at three points, and F (respectively,, and ) are chosen so that the quadrilateral F is a square. If x is the length of the side of the square, show that 1 x =

5 Randolph High School Math League Page [HSM 1981] In, M is the midpoint of side, N bisects, and N N. If sides and have lengths 14 and 19, respectively, then find MN. (Hint: xtend N past N to intersect at a point Q.) 14 N M [Mandelbrot ] pyramid has a square base, triangular sides, and eight edges that are each 80 meters long. straight path begins at one corner of the square base, slanting upwards to meet the next edge at a point 0 meters along that edge from the corner, as shown. The path continues around the pyramid, always slanting upward at the same angle, making infinitely many turns. What is the total length of the path? 1. [OMO 2014] The points,,,, lie on a line l in this order. Suppose T is a point not on l such that T = T, and T is tangent to the circumcircle of triangle T. If = 2, = 6, and = 15, compute. 14. [Math League HS ] What is the area of a trapezoid whose altitude has a length of 12 and one of whose perpendicular diagonals has a length of 15? 6. Problem Set 15. [IM 1998] Let be a parallelogram. xtend through to a point P, and let P meet at Q and at R. Given that P Q = 75 and QR = 112, find R. 16. [Thomas Mildorf] is an isosceles triangle with base. is a point on and is the point on the extension of past such that is right. If = 15, = 2, and = 16, then can be expressed as m n, where m and n are relatively prime positive integers. etermine m + n. 17. [Math League HS ] In, = 18, and is the point on for which = 5. Perpendiculars drawn from to and have lengths 4 and 5 respectively. What is the area of? 18. [IM 1986] In, = 425, = 450, and = 510. n interior point P is then drawn, and segments are drawn through P parallel to the sides of the triangle. If these three segments are of an equal length d, find d.