# SMT 2018 Geometry Test Solutions February 17, 2018

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2 SMT 018 Geometry Test Solutions February 17, 018 Let x denote the length of a a. Since D 1 is a 1-gon, each angle is = 150 degrees. We know that a 11 a a 5 is the right angle of a square, and by symmetry a 11 a a 1 = a 5 a a, so Xa a = = 0. Again, by symmetry we also have Xa a = 0. Using right triangles, we see that a X = x/. On the other hand, since Za 1 a = 0 by symmetry and a 1 Za = Y ZX = 0 by similar triangles, we see that a 1 Za is isoceles, and therefore a Z = x. Since XZ = a Z a X, we have XZ = x x/. Therefore, XZ a X = x x/ x/ = 1. Squaring this gives us the ratio of areas In ABC, ABC = 75 and BAC is obtuse. Points D and E are on AC and BC, respectively, such that AB BC = DE EC and DEC = EDC. Compute DEC in degrees. Answer: 85 Solution: Extend AC past A, and draw F on AC such that AB = F B. Note that AB F B, and since F BC = DEC we have F BC DEC. BC = DE EC BC = Next, we perform some angle chasing. Let DEC = EDC = x. By similar triangles, we have F BC = BF C = x as well. Furthermore, F B = AB, so F AB is isoceles, and thus F AB = x as well. Now ABC = 75, so we compute F BA = F BC ABC = x 75. The angles of a triangle sum to 180, giving us the equation x 75 = 180, which solves to x = 85.. In ABC, AB =, AC =, and D is drawn on BC such that AD is the angle bisector of BAC. D is reflected across AB to a point E, and suppose that AC and BE are parallel. Compute CE. Answer: 1 Solution: Let BAC = x. Since AC BE, we have ABE = x, so ABC = x since E is the reflection of D across AB. This means that ABC is isosceles, so AC = BC =. Using the angle bisector theorem, we find CD = 4 and BD = BE =. Now let θ = CBA. We can use the Law of Cosines to compute CE = BC + BE BC BE cos θ. Because ABC is isoceles, we can drop the perpendicular from C to find cos θ = 1.5 = 1 4. Using the double angle formula, we get cos θ = cos θ 1 = 7 8. Plugging this in gives us the answer ( ) 7 CE = + = Two equilateral triangles ABC and DEF, each with side length 1, are drawn in parallel planes such that when one plane is projected onto the other, the vertices of the triangles form a regular hexagon AF BDCE. Line segments AE, AF, BF, BD, CD, and CE are drawn, and suppose that each of these segments also has length 1. Compute the volume of the resulting solid that is formed. Answer: Solution 1: Draw lines l 1, l, l through A parallel to BC, through B parallel to CA, and through C parallel to AB, respectively. Suppose that l and l intersect at D, l and l 1 intersect at E, and l 1 and l intersect at F. We observe that DD, EE, and F F are concurrent at a single point G. In addition, D E F is an equilateral triangle with side length, double that of DEF. Let the notation [f] denote the volume of a figure f. Then, we observe that [D E F G] = [ABCDEF ] + [ABF F ] + [BCDD ] + [CAEE ] + [DEF G]. Next, we wish to show that

3 SMT 018 Geometry Test Solutions February 17, 018 DD = EE = F F = 1; this would allow us to show that ABF F, BCDD, CAEE, DEF G are regular tetrahedra with side length 1, and that D E F G is a regular tetrahedra with side length. Consider the trapezoid EE F F. Assume for the sake of contradiction that EE 1. Then F F 1 as well, since EF E F. Since AE = AE = AF = AF = 1, both AEE and AF F are not equilateral, and EAE = F AF is not 0. However, this means that EAF 0 since E AE + EAF + F AF = 180, contradicting the fact that AEF is equilateral with side length 1. Hence, we have DD = EE = F F, after extrapolating the previous argument to the third side. Therefore, [ABCDEF ] = [D E F G] 4[DEF G] = 8[DEF G] 4[DEF G] = 4[DEF G], since a volume which is scaled by twice the side length has its volume scaled by = 8. It suffices to compute the volume of a regular tetrahedron. Dropping the altitude from G to DEF, we can compute the height of the tetrahedron DEF G to be of DEF G is Finally, our desired area is 4 1 A[ DEF ] = 1 =. 4 = 1.. Hence, the volume Solution : Because each of the edges of the solid have length 1, each of the faces of the solid are equilateral triangles. There are 1 edges, which implies that the solid is in fact a regular octahedron with edge length 1. To compute the volume of a regular octahedron, we split it into two congruent square pyramids. Let ABCD be the square base with side length 1, and let E be the top of the pyramid. Also, let F be the point in the middle of ABCD and let M be the midpoint of AB. Clearly the base ABCD has area 1. To compute the height EF, we note that EF M is a right triangle with F M = 1. To compute EM, we note that it is a leg of the right triangle AME where AM = 1 and AE = 1. Using the Pythagorean Theorem on AME gives us EM =, and using it again on EF M gives us the height EF =. Therefore, the volume of the square pyramid ABCDE is 1 1 the regular octahedron is. =, which implies that the volume of 8. Let ABC be a right triangle with ACB = 0, BC = 1, and AC = 1. Let the angle bisectors of BAC and ABC intersect BC and AC at D and E respectively. Let AD and BE intersect at I, and let the circle centered at I passing through C intersect AB at P and Q such that AQ < AP. Compute the area of quadrilateral DP QE. Answer: 1 Solution: We claim that AP = AC. Let P be a point on AB such that AP = AC. Since AI = AI and CAI = P AI, we have ACI = AP I by SAS congruency. Thus, IC = IP, so P must lie on the circle centered at I passing through C. This implies that either P = P or Q = P. Let the circle centered at I passing through C intersect AC at X. Note that AX < AC, and by Power of a Point, AX AC = AQ AP. Now suppose that P = Q. Then AC = AP = AQ, so AP = AX < AC = AQ, contradicting AQ < AP. Hence, P Q, so P = P and AP = AC.

5 SMT 018 Geometry Test Solutions February 17, Let ABC be a triangle with AB = 1, AC = 14, and BC = 15, and let Γ be its incircle with incenter I. Let D and E be the points of tangency between Γ and BC and AC respectively, and let ω be the circle inscribed in CDIE. If Q is the intersection point between Γ and ω and P is the intersection point between CQ and ω, compute the length of P Q. Answer: 8 Solution: We can derive that CD = CE = 8. We then compute the inradius r of ABC. Using Heron s Formula or drawing an altitude from B to AC, we can calculate that the area of ABC is 84. Since the product of r and the semiperimeter of ABC also gives the area, we find that r = 4. Let O be the center of ω. Also let ω touch ID at X and CD at Y. Since OXDY is a square, we have IXO OY C. Let x be the radius of ω, giving us XO = Y O = x and IX = 4 x and CY = 8 x. This gives us the equation 4 x x = x 8 x, and solving for x yields x = 8. Since both ω and Γ are tangent to AC and BC, a homethety (the enlargement/shrinking of objects with respect to a fixed point and fixed ratio) centered at C sends ω to Γ, and the ratio is x r =. Since C, P, and Q are collinear, the same homothety also takes P to Q, so CP CQ =. Letting CP = k and CQ = k, we have that P Q = k. Finally, CY = 8 8 = 1, so by Power of a Point, CY = CP CQ = ( ) 1 = k. Solving for k gives us P Q = 8.

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