12 th Annual Johns Hopkins Math Tournament Saturday, February 19, 2011

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1 1 th Annual Johns Hopkins Math Tournament Saturay, February 19, 011 Geometry Subject Test 1. [105] Let D x,y enote the half-isk of raius 1 with its curve bounary externally tangent to the unit circle at the point (x, y, such that the straight bounary of the isk is parallel to the tangent line (so the point of tangency is the mile of the curve bounary. Fin the area of the union of the D x,y over all (x, y with x + y = 1 (that is, (x, y is on the unit circle. 4π Given any point of tangency (x, y, the points on the circle farthest away are the two iametrically opposite points on the straight bounary. The length of the segment connecting (x, y an either of these two points (pick one an call it A is because the half-isk has raius 1; the line segment can be viewe as the hypotenuse of the right triangle spanne by two raii of D x,y. As the union of the D x,y sweep aroun the unit circle, the segment from (x, y to A sweeps out an annulus of with ; that is, it sweeps out a large circle of raius 5 with an inner circle of raius 1 remove. Hence the area of the remaining region is ifference between the areas of the two circles: 5π π = 4π.. [106] Let circle O have raius 5 with iameter AE. Point F is outsie circle O such that lines AF an EF intersect circle O at points B, an D, respectively. If AF = 10 an m F AE = 0, then the perimeter of quarilateral ABDE can be expresse as a + b + c + 6, where a, b, c, are rational. Fin a + b + c After some angle chasing, we fin that m DBF = m DF B = 75, which implies that DF = DB. Hence the esire perimeter is equal to AF BF + AE + F E = 0 BF + F E. By the Law of Sines, F E sin 0 = 10 sin 75 F E = = Now, to fin BF raw the altitue from O to AB intersecting AB at P. This forms a triangle, so we can see that AP = 5 = 10 BF BF = Hence the answer is 0 + (5 6 5 (10 5 = , so the esire sum is = 15.. [108] In a unit cube ABCD EF GH, an equilateral triangle BDG cuts out a circle from the circumsphere of the cube. Fin the area of the circle. π Consier the cube of sie length an ivie the answer by 4 later. Set the coorinates 6 1

2 of the vertices of the cube to be (±1, ±1, ±1. Then the plane going through a equilateral triangle can be escribe as x + y + z = 1. The istance to the plane from the origin is 1, as ( 1, 1, 1 is the foot ( of the perpenicular from (0, 0, 0. Thus the raius of the circle is 1 1 =, so the area is π. In the case of the unit cube we shoul ivie it by 4 to get the answer π [10] Compute the largest value of r such that three non-overlapping circles of raius r can be inscribe in a unit square The three circles will be inscribe in such a way that one altitue of the equilateral triangle forme by the centers of the three circles will coincie with a iagonal of the square, as in the figure below. Inee, by the Pigeonhole Principle, one circle must lie tangent to two sies of the square, an in any orientation other than the one below, the circles can be ilate. Label the square ABCD starting in the upper left an going clockwise. The line from the center of the top left circle to A has length r, an the equilateral triangle forme by the raii has height r. Then the line from the base of the equilateral triangle to C has length ( + r. Now, raw lines from the centers of the two lower circles to C to form four triangles. Observe that these triangles are ientical, with angle π 8 in the lower right. Then ( π tan = 1 = 8 r ( + r an solving for r gives the esire answer, or some equivalent expression. 5. [1040] Let ABCD be a unit square. Point E is on BC, point F is on DC, AEF is equilateral, an GHIJ is a square in AEF such that GH is on EF. Compute the area of square GHIJ The setup is as follows: First let a be the length of AE. Then CE = a/, BE = 1 a/ so AE = a = 1 + BE = a + a /. Solving it gives a + a 4 = 0, (a + = 6 so a = 6. Next let b

3 be the length of IJ. Then AIJ is equilateral so AJ = b. Also JE = / b, so AE = a = + b, b = ( ( ( 6 = (9 5. Squaring it gives [1056] Let ABC be equilateral. Two points D an E are on sie BC (with orer B, D, E, C, an satisfy DAE = 0. If BD = an CE =, what is BC? A 0 B D E C Rotate the figure aroun A by 60 so that C comes at the place of B. Let B, C, D, E be corresponing points of the move figure. Since E AD = E AC + C AD = EAC + BAD = 0 = EAD, E A = EA an DA = D A, one has E D = ED. So BC = BD + DE + EA can be foun out if we know E D. But E D = E B + BD E B BD cos 10 = 19, so BC = = B D A E B = C D E C 7. [1088] Let ABCD be cyclic quarilateral with AB = 6, BC = 1, CD =, an DA = 6. Let E, F be the intersection of lines AB an CD, lines AD an BC, respectively. Fin EF. 10 We have ADE CBE, an ratio is AD : CB = 1 :. Let AE = p an DE = q, then we have AB = BE AE = DE AE = q p an CD = p q. Solving for p an q we have p = 4 an q = 5. Similarly we have F C = 8 an F D = 10.

4 Let B = θ, then F DE = π θ. Apply the cosine law to EBF to get EF = BE + BF BE BF cos θ = cos θ = cos θ an to EDF to get EF = DE + DF + DE DF cos θ = cos θ = cos θ. Solving for EF we get EF = [115] Two parallel lines l 1 an l are on a plane with istance. On l 1 there are infinitely many points A 1, A, A, progressing in the same istance: A n A n+1 = for all n. In aition, on l there are also infinitely many points B 1, B, B, satisfying B n B n+1 = 1 for all n. Given that A 1 B 1 is perpenicular to both l 1 an l, express the sum i=1 A ib i A i+1 in terms of. π tan 1 ( 1 Construct points C 1, C, C,... on l 1 progressing in the same irection as the A i such that C 1 = A 1 an C n C n+1 = 1. Thus we have C 1 = A 1, C = A, C 5 = A, etc., with C n 1 = A n in general. We can write A i B i A i+1 = C i 1 B i C i+1 = C i B i C i+1 C i B i C i 1. Observe that C i B i C k (for any k is a right triangle with legs of length an k i, an C i B i C k = tan 1 ( k i. So Ci B i C i+1 C i B i C i 1 = tan 1 ( i+1 which has nth partial sum so it converges to π tan 1 ( 1. i=1 ( tan 1 i 1 ( ( i + 1 i 1 tan 1 tan 1 ( n + 1 tan 1 ( + tan 1 n ( 1 tan 1. The whole sum is therefore 9. [180] In an unit square ABCD, fin the minimum of AP + BP + CP when P is an arbitrary point in ABCD. 5 Rotate the triangle AP B aroun A by 90 egree as in the following figure. Let P an B be the rotate image of P an B respectively. Then we have B P = BP, P P = AP so AP + BP + CP = CP = P P + P B CB = 5. 4

5 10. [156] Given a triangle ABC with sie lengths a = 5, b = 7, c = 8, fin the sie length of largest equilateral triangle P QR such that A, B, C are on QR, RP, P Q, respectively. Answer 4 We claim that in general, the answer is (a + b + c + 4 S, where S is the area of ABC. Suppose that P QR is a equilateral triangle satisfying the conitions. Then BP C = CQA = ARB = 60. The locus of points satisfying BXC = 60 is an part of a circle O a. Draw O b an O c similarly. Three circles meet at a single point X insie the triangle, which is the unique point satisfying BXC = CXA = AXB = 10. Then the choice of P on O a etermines Q an R: those two points shoul also be on O b an O c respectively, an P CQ an P BR shoul form the sie of triangle. Now one shoul fin the maximum of P Q uner these conitions. Note that BP X an BRX oes not epen on choice of P, so triangle P XR has same shape. Especially, the ratio of P X an P R is constant, so P R is maximize when P X is the iameter of O a. This requires P Q, QR, RP to be perpenicular to XC, XA, XB respectively. From this point there may be several ways to calculate the answer. One way is to observe P Q = (AX + BX + CX by consiering (P QR = (P XQ + (QXR + (RXP. AX + BX + CX can be compute by the usual rotation trick for the Fermat point: rotate BXA 60 aroun B to BX A. Observe that BXX is equilateral, an so A, X, X, an C are collinear. Hence, A C = AX +BX + CX, an we can apply the Law of Cosines to A BC to get that A C = c + a ac cos (B + 60 = a + c + ac sin 60 sin B ac cos 60 cos B = a + c + S 1 (a + c b = a + b + c + S = P Q = (a + b + c + 4 S (where S is again the area of ABC. Plugging in our values for a, b, an c, an using Heron s formula to fin S = 10 5 = 10, we can calculate P Q = 4. 5

SMT 2018 Geometry Test Solutions February 17, 2018

SMT 2018 Geometry Test Solutions February 17, 2018 SMT 018 Geometry Test Solutions February 17, 018 1. Consider a semi-circle with diameter AB. Let points C and D be on diameter AB such that CD forms the base of a square inscribed in the semicircle. Given