11 th Annual HarvardMIT Mathematics Tournament


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1 11 th nnual HarvardMIT Mathematics Tournament Saturday February 008 Individual Round: Geometry Test 1. [] How many different values can take, where,, are distinct vertices of a cube? nswer: 5. In a unit cube, there are types of triangles, with side lengths (1, 1, ), (1,, ) and (,, ). Together they generate 5 different angle values.. [] Let be an equilateral triangle. Let Ω be its incircle (circle inscribed in the triangle) and let ω be a circle tangent externally to Ω as well as to sides and. etermine the ratio of the radius of Ω to the radius of ω. nswer: Label the diagram as shown below, where Ω and ω also denote the center of the corresponding circles. Note that M is a median and Ω is the centroid of the equilateral triangle. So M = MΩ. Since MΩ = NΩ, it follows that M/N =, and triangle is the image of triangle after a scaling by a factor of, and so the two incircles must also be related by a scale factor of. ω N Ω M. [4] Let be a triangle with = 90. circle is tangent to the sides and at X and Y respectively, such that the points on the circle diametrically opposite X and Y both lie on the side. Given that = 6, find the area of the portion of the circle that lies outside the triangle. Y X nswer: π Let O be the center of the circle, and r its radius, and let X and Y be the points diametrically opposite X and Y, respectively. We have OX = OY = r, and X OY = 90. Since triangles X OY and are similar, we see that =. Let X be the projection of Y onto. Since X Y is similar to, and X Y = r, we have X = r. It follows that = r, so r =. X Y O Y X X 1
2 Then, the desired area is the area of the quarter circle minus that of the triangle X OY. nd the answer is 1 4 πr 1 r = π. 4. [4] In a triangle, take point on such that = 14, = 1, = 4, and the circumcircle of is congruent to the circumcircle of. What is the area of triangle? nswer: 108 M The fact that the two circumcircles are congruent means that the chord must subtend the same angle in both circles. That is, =, so is isosceles. rop the perpendicular M from to ; we know M = 9 and so M = 5 and by Pythagoras on M, M = 1. Therefore, the area of is 1 (M)() = 1 (1)(18) = [5] piece of paper is folded in half. second fold is made such that the angle marked below has measure φ (0 < φ < 90 ), and a cut is made as shown below. Á When the piece of paper is unfolded, the resulting hole is a polygon. Let O be one of its vertices. Suppose that all the other vertices of the hole lie on a circle centered at O, and also that XOY = 144, where X and Y are the the vertices of the hole adjacent to O. Find the value(s) of φ (in degrees). nswer: 81 Try actually folding a piece of paper. We see that the cut out area is a kite, as shown below. The fold was made on, and then and. Since was folded onto, we have =. ither or is the center of the circle. If it s, then = 144, so = 7. Using =, we see that = = 54. So = 7, and thus φ = = 99, which is inadmissible, as φ < 90. So is the center of the circle. Then, = = 54, = 7, and φ = = [5] Let be a triangle with = 45. Let P be a point on side with P = and P = 5. Let O be the circumcenter of. etermine the length OP.
3 nswer: 17 Using extended Sine law, we find the circumradius of to be R = sin = 4. y considering the power of point P, we find that R OP = P P = 15. So OP = R 15 = = [6] Let 1 and be externally tangent circles with radius and, respectively. Let be a circle internally tangent to both 1 and at points and, respectively. The tangents to at and meet at T, and T = 4. etermine the radius of. nswer: 8 Let be the point of tangency between 1 and. We see that T is the radical center of the three circles, and so it must lie on the radical axis of 1 and, which happens to be their common tangent T. So T = 4. 1 T We have tan T = T = 1 T, and tan Thus, the radius of equals to T tan T ( ) T + T = 4 tan tan T + tan T = 4 1 tan T = 4 = tan T = T = [6] Let be an equilateral triangle with side length, and let Γ be a circle with radius 1 centered at the center of the equilateral triangle. etermine the length of the shortest path that starts somewhere on Γ, visits all three sides of, and ends somewhere on Γ (not necessarily at the starting point). xpress your answer in the form of p q, where p and q are rational numbers written as reduced fractions. 8 nswer: 1 Suppose that the path visits sides,, in this order. onstruct points,, so that is the reflection of across, is the reflection of across, and is the reflection of across. Finally, let Γ be the circle with radius 1 centered at the center of. Note that Γ is the image of Γ after the three reflections:,,.
4 When the path hits, let us reflect the rest of the path across and follow this reflected path. When we hit, let us reflect the rest of the path across, and follow the new path. nd when we hit, reflect the rest of the path across and follow the new path. We must eventually end up at Γ. It is easy to see that the shortest path connecting some point on Γ to some point on Γ lies on the line connecting the centers of the two circles. We can easily find the distance between the two centers ( ) to be + 1 = 8. Therefore, the length of the shortest path connecting Γ to Γ has length 8 1. y reflecting this path three times back into, we get a path that satisfies our conditions. 9. [7] Let be a triangle, and I its incenter. Let the incircle of touch side at, and let lines I and I meet the circle with diameter I at points P and Q, respectively. Given I = 6, I = 5, I =, determine the value of (P/Q). nswer: Q F I P Let the incircle touch sides and at and F respectively. Note that and F both lie on the circle with diameter I since I = F I = 90. The key observation is that,, P are collinear. To prove this, suppose that P lies outside the triangle (the other case is analogous), then P = P I = I+ I = 1 ( + ) = 90 1 =, which implies that,, P are collinear. Similarly, F, Q are collinear. Then, by Power of a Point, P = F Q. So P/Q = F/. Now we compute F/. Note that F = sin I = 6 ( 6) =, and = sin I = 5 ( ) 5 = 4 5. Therefore, F/ = [7] Let be a triangle with = 007, = 008, = 009. Let ω be an excircle of that touches the line segment at, and touches extensions of lines and at and F, respectively (so that lies on segment and lies on segment F ). Let O be the center of ω. Let l be the line through O perpendicular to. Let l meet line F at G. ompute the length G. nswer: Let line meet ω again at H. Since F and are tangents to ω and H is a secant, we see that HF is a harmonic quadrilateral. This implies that the pole of with respect to ω lies on F. Since l, the pole of lies on l. It follows that the pole of is G. 4
5 G F O H Thus, G must lie on the tangent to ω at, so,,, G are collinear. Furthermore, since the pencil of lines (, F ;, G) is harmonic, by intersecting it with the line, we see that (, ;, G) is harmonic as well. This means that G G = 1. (where the lengths are directed.) The semiperimeter of is s = 1 ( ) = 01. So = s 009 = 100 and = s 008 = Let x = G, then the above equations gives Solving gives x = x x 100 = 1. Remark: If you are interested to learn about projective geometry, check out the last chapter of Geometry Revisited by oxeter and Greitzer or Geometric Transformations III by Yaglom. 5
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