GEOMETRY TEAM February 2018 Florida Invitational. and convex pentagon KLMNP. The following angle

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1 February 0 Flrida Invitatinal Cnsider cnvex hexagn EFGHIJ measures, in degrees are interir angles t these plygns: me 00, mf 0, mg x 60, mh x 0, mi 0 mk 0, ml 00, mm 90, mn y 90, m y and cnvex pentagn KLMN The fllwing angle, mj 0 A = the value f x B = the value f C = the degree measure f the largest exterir angle f hexagn EFGHIJ D = the degree measure f the largest interir angle f pentagn KLMN y QS and URT are drawn s that Q RU and UT S RU=VT=, QV=QT=6, V= and the pints R, Q, T and S are cllinear at V Q intersects TU U may nt be drawn t scale A = the length f QR B = the length f S C = the area f TUR, in fractin frm 6 D = the number f distinct angles, SHOWN in the diagram, which are cngruent t 6 R Q T S Cnsider the fllwing pints n the xy-plane: (, 0), Q (, ) and R( k, k ) A = the length f segment Q B = n k if the midpint f R has crdinates (6, n ) C = k if the slpe f the line perpendicular t the line thrugh pints Q and R has the same slpe as the line thrugh pints and Q Give C in fractin frm D = the value f k if the pint R is equidistant frm and Q Give D in fractin frm V

2 February 0 Flrida Invitatinal A = the area f a circle with circumference B = the circumference f a circle with area R C: In the figure t the right, each f fur cngruent circles are internally tangent t the sides f a 6x6 square 6 and t tw ther circles is a pint f tangency f ne circle t the square, and R is a pint f tangency f the tw tp circles, as shwn, and Q is a vertex f the square Give the perimeter f Q 6 D: Give the area f the square which is utside f all fur circles (shaded) A gat (G) is n a leash attached t the grund at the midpint M f a rectangular pen that is 0 feet by 9 feet The leash is feet lng There is a break in the pen s vertex n the vertex near M as shwn, and it is just large enugh fr the gat t get thrugh 0 Cnsider the dimensins f the break and f the gat t be negligible (disregard them in yur cmputatins) 9 RQ G A = the area in square feet that the gat can ram utside f the pen Assume the leash, gat and M grund are all cplanar B = if the gat can ram inside f the pen an area f 0 sq ft B = 0 if the gat cannt ram inside f the pen an area f 0 sq ft Assume the leash, gat and grund are all cplanar C = the amunt f square feet that the gat wuld be able t ram inside the pen if the leash were 9 feet lng Assume the leash, gat and grund are all cplanar D: The gat heads tward the tp right vertex () f the pen He stps when the leash is taut, and he is as clse as he can get t The distance in feet at that spt frm the 0-ft side f the pen is a b b, fr b a psitive prime number Give the value f a b Assume the leash, gat and grund are all cplanar E F J K N break 6 figure figure figure H G L M R Q EFGH is a JKML is NQR is a trapezid parallelgram a rhmbus with bases N and RQ A = the value f x y, if me (x y), mh 6, and mf y (figure abve) B = the perimeter f JKML if JM=6 and LK=0 (figure abve) C = m LJK if mljk mjkl 5 (figure abve) D = the area f NQR if NR=Q=, mn 60 and RQ= (figure abve)

3 February 0 Flrida Invitatinal 7 A = the area f the circle inscribed in a regular hexagn with perimeter B = the area f the circle inscribed in an equilateral triangle with area 6, in fractin frm C = the area f the square circumscribed abut a circle which is circumscribed abut an equilateral triangle with perimeter D = the ttal circumference f the three tangent circles E shwn t the right, if the centers f the circles are E, F and G, and EFG has perimeter 00 F G SHOWN IN ICTURE: G is n FH which is parallel t JL and M J, K and L are cllinear M, N and are cllinear Transversal EQ intersects the parallel lines at G, K and N as shwn Angle measures in degrees are x y, y 5 and x 0 fr angles FGK, GKJ and KN respectively (nt in kids picture) THE DIAGRAM IS NOT TO SCALE A = the value f x B= the value f C = mfgk m GKJ D = the distance between lines FH and M, if GK=0, KN=, and lines FH and JL are a distance f apart lygn is regular and has ne interir angle which is 7 degrees lygn lygn y is regular and has 09 diagnals is regular and has interir angles which ttal 700 degrees lygn is regular and has fur mre sides than plygn A cm = the perimeter f plygn if ne side f x y J y 5 K L has length cm x 0 B cm = the perimeter f plygn if ne side f has length k cm and has (k+) sides C = the degree measure f ne interir angle f plygn Write C in fractin frm D = the degree measure f ne exterir angle f plygn Write D in fractin frm E F G H M N Q

4 February 0 Flrida Invitatinal 0 Square S (fr small square) is inscribed in circle Square L (fr large square) is circumscribed abut circle A = the perimeter f square L if the perimeter f square S is 0 B = the area f square S if the area f square L is 0 C = the value f C s that the rati f the areas f circle C t square S is Yur answer shuld be in fractin frm and shuld include n square rt D = the rati f the areas f square L t circle Tangent segment Q is part f tangent line Q with the pint f tangency Q, R, S and U are cllinear Q, S and T are cllinear ints R, Q T and U are n the circle Q=6 RS= TS= Remember: diagrams may nt be drawn t scale! 6 Q A = the length f SU B = the length f SQ C: Let C = if QS is acute, let C= if QS is btuse, and let C= if QS is a right triangle D: is a diameter f the circle If W=0 then give the circumference f the circle EFG has side lengths 9, and 5 Right HIJ has tw sides with lengths 5 and Right QR has altitude t hyptenuse Q with length 0, and QR with length 5 QW A = the length f the altitude t the lngest side f B = the least pssible length f a side f HIJ C = the length f hyptenuse Q in QR D = the least pssible length f an altitude t HIJ EFG R S T U

5 U GEOMETRY TEAM February 0 Flrida Invitatinal T L E ( p ) (p 50) R S Y (n 0) (n 50) X 0 Q Triangle TRU is issceles with TR=UR Triangle YXQ has tw sides mt p mtrs (p 50) extended as shwn, s Y, X and E U, R and S are cllinear cllinear, and Q, X and L cllinear my ( n 0) mlxe (n 50) mq 0 (D it bth ways) A = the value f B = the value f n C = the area f a square with ne diagnal length ( ) (Yur answer shuld cntain n variables) (D it bth ways) p n p D = the area f a regular ddecagn ( sides) with side length n, using tan(5) = (Yur answer shuld cntain n variables) (D it bth ways) n 5 R Tw equilateral triangles RYN and SV verlap s that each f six small triangles (ne is RTU) are cngruent The hexagn in the center is regular RYN and SV each have sides f length S W T U V X A = the area f the star with pints R, V, N,, Y and S, that is, the area f the nn-cnvex ddecagn B = the area f the hexagn TUXMZW C = the perimeter f the star described in part A D = the distance R, which is the distance between ppsite pints f the star Y Z M N

6 February 0 Flrida Invitatinal ANSWERS art A art B art C art D Ntes: r r r 05 5 art C: Radical may be in numeratr Fractin frm needed r 7 7 r r art C, D must be fractin frm art C can be in any rder f terms art B must be in fractin frm arts C and D must be in fractin frm Nte there is n r 75 r 75 r 70 r in parts C and D C must be fractin frm

7 VQT QVT GEOMETRY TEAM February 0 Flrida Invitatinal TEAM SOLUTIONS Hexagns have angles that ttal t (6 )0 70 degrees The ttal fr given angles are 00+0+x+60+x = 50+5x entagns have angles that ttal t (5 )0 50 Ttal fr the given pentagn angles are y+90+y= 90+y art A: 5x+50=70 x=90/5= A= art B: y+90=50 y=50 B=50 art C: The largest exterir is a linear pair with the smallest interir Angle E is 00, which is smallest, s the exterir at E must be 0 art D: The largest interir is y+90= 0 may nt be A: Since RUT and are issceles and similar, drawn t RT=, and since QT=6, QR= A= scale B: Cmpare similar triangles VQT and QS U VQ Q 6 9 S= V VT S S V C: It is easier t get the height drawn (t the 6 right) f 6 6 X and use similarity ythagrean Th 6 says R Q T S Q 6 T QX= 6 6 = 5 Area f QVT is 5 5 The rati f TUR and VQT is 6/ r ¾ S areas are in the rati f 9/ 9 5 S the area f TUR is 5 6 areatur 9 D: The base angles f the three similar triangles are cngruent QVT, VTQ, U,, S D=5 angles ints given are (, 0), Q (, ) and R( k, k ) A: Q= ( ) ( 0) = 9 5 k 0 k B: the midpint f R is, = (6, n ) k frm x crdinates k=9 If k=9, y-values give 0 9 n n=/ nk 95 r 9/ k k C: Q and R has slpe and Q has slpe Since they are k k 9 k perpendicular, k6 k k= 6 k 7 D: ( k ) ( k ( 0)) ( k ) ( k ) k 6k 9 ( k ) ( k ) k The tw quantities have nly the middle term different when expanded 6k 9k k k = -9 k

8 art A: r art B: r 6 r GEOMETRY TEAM February 0 Flrida Invitatinal Area is r 7 7 r Circumference is = art C: RQ= because the right triangle shwn with R as R hyptenuse has sides f twice a radius each RQ is the hyptenuse f a triangle with legs and, s RQ = Q is the hyptenuse f a triangle with legs and 6, and has length 7 erimeter Q 6 is art D: Each circle has area 6 s area f the shaded regin is art A: He has ft f leash and needs 0 ft f it t get t the break in the pen S he has feet f leash t rtate abut the vertex f the break He can ram ¾ f a circle f radius, s area is art B: He cannt ram all f it! He has enugh leash t cver all f the 0 ft side Cnnect the leash tether pint t a tp vertex That is His leash is His leash will nt reach the tp vertices S B=0 art C: If the leash has length 9 then the leash will reach t the tp f the pen, perpendicular t the 0-ft side He can ram a semi-circle inside, radius 9 r 05 art D: x 0 x = 9 0 = a b b ab 9 6 x E F J 7 K N 7 7 x-y y 60 5 figure figure figure 6 H G L M R Q A: y=6 s y= x-y=6 x-=6 x=0 x=0 x+y= B: Since the diagnals f a rhmbus are perpendicular, we can use knwledge f Triples, r the ythagrean Therem t get ne side f rhmbus JKML is and s perimeter f the rhmbus is 7()=6 C: We knw that ne diagnal will bisect an angle f a rhmbus, s let the btuse angle at J be n Then the acute angle at K will be 0-n S we say n 0 n 5 as we are given mljk mjkl 5 S duble t get n0 n 90 n 0 = m J C=0 D: Drp tw altitudes frm the small base t the larger base in trapezid NQR The height is 7 since Angle N is 60 degrees See figure abve Area =

9 GEOMETRY TEAM February 0 Flrida Invitatinal 7 = 05 7 A: Drp the apthem and draw the radius f the hexagn and we get a triangle as shw t the left (ne angle f a regular hexagn is 60 degrees The circle then has radius and the area f the circle is side B: 6 slves t side= f the triangle Again drp an apthem which is the radius f the circle The lng leg f the triangle is half the length f the triangle side, Apthem and area f the circle is 6 C: Triangle side is 6 Half is, which is ne side f the triangle t the left Divide by t get the triangle apthem is and the triangle (and circle) radius is The circle radius is half f the square side, s the square side is Area f the square is 6()= D: If the circle radii are a, b and c, then the perimeter f triangle EFG is a+b+c=00 Since ne circumference E is r times pi, we have a b c 00 and a a c s the sum f the circumferences is 00 b art A: Using Alternate Exterir Angle Th and b c Cnsecutive Interir Angle Th, x y x 0 and F G x y y 5 0 xy 0 --> x y 5 xy 5 Subtract the last tw equatins, y 0 y=5 Substitute x-5=5 x=50 A=50 art B: B=5 (see part A) art C: mfgk m GKJ = [50+(5)]-[5-5]=0-0=00 GK 0 5 art D: KN = and M is + = r 56 5 FH d 0 / 6 5d d=distance JL t M= Distance between 9 art A: Subtract frm 0 t get exterir angle, 0-7=9 deg 60/9=0 sides erimeter = (0)=60 art B: Trial and errr wrks, r nn ( ) 09 nn ( ) (09)=()(9) S n= and (n-)=9 sides = k+ k=0 Side length is k erimeter is 0()=0 art C: 0( n ) /0=70/=0/=0=n- n= sides One interir angle is 700/ = 600/= 00/7 art D: lygn has sides, s lygn has 6 sides

10 February 0 Flrida Invitatinal 0 art A: Small perimeter is 0 Small side is 0 Diagnal f S is 0 Radius f circle is 5 Square L has side equal t the diameter f the circle, 0 erimeter f L is (Nte rati f the sides is : ) art B: Large square has area 0 and side 0 Circle diameter is als 0 This 0 is diagnal f the small square, s the side is 5 Area is 0 (Since rati f the sides is :, rati f areas is :) art C: Use values frm part A fr cnvenience: circle has radius 5 and area Square S has area 00 Rati is 50 :00 : The rati is 50 C C Square bth sides: C C= art D: Again, using part A values fr cnvenience, square L has area 00 and the circle (see part C) has area 50 Rati is D s D= art A: Using the exterir secant segment frmula, (5 SU ) 6(6), 5+SU=6 SU= SU=7 art B: Using infrmatin fund in part A, (7)=(SQ) SQ= 7/ art C: Using infrmatin fund in part B, QS has side lengths 5, 6 and 7/ Using the smallest tw, If the triangle were right, this wuld be the length f the hyptenuse The third side is 6, which is S, the triangle is acute, and C= art D: Lking at the small diagram t the right, W= Q All three triangles given are right triangles = 6 Q R S 7 which is smaller art A: Area f EFG is (9)() 5 by using legs as base and height Nw change t cnsider the hyptenuse the base, s (5) h r 7 5 art B: The least length f the missing side will ccur if the greatest side is Then the least pssible side length fr the missing side will be Nw W cntenders fr the smallest side are 5 and 9 Since 5 5, this is smallest 0 T U

11 February 0 Flrida Invitatinal art C: Let S be the pint where Q intersects the altitude t Q Use the ythagrean Therem t get QS= Then use Gemetric Mean t get ( S) S= Q= = 9 5 art D: Case : In HIJ, we have sides 5,, and 5 6 9, The shrtest altitude in a triangle will be t the lngest side Area is h Case : Sides are 5, Cmpare: dividing and see and 9, Area is = h = () Height wuld be h 5(9) 6 is smaller, s the answer is = Nw we cmpare by art A: The measure f an exterir angle is equal t that f the sum f the remte interir angles Since TR-UR, the remtes are equal, s pp 50 p=0 art B: Using the angle vertical t <LXE, n 0 0 n n=70 n=r 75 art C: If the diagnal is p+n=, then the side is, and the area is ()= Or 75 art D: The angle created by a radius f the ddecagn and the apthem is 60/ = 5 degrees 7 tan 5 apthem = 6 a 5 Area = ( apthem) perimeter 6 = ()()= r 70 Each small equilateral triangle must have sides f length art A: The star is hexagn plus 6 triangles () 6 = art B: See part A: art C: Six pints, each pint with tw sides, gives ()= Y art D: This is tw equilateral triangle altitudes, plustz which is Z M N the lng side f an issceles triangle with sides and and included angle 0 The altitude f the triangle is Lng side f the issceles triangle described is Final distance is 7 Q 5 S 0 R 5 a S W R T U V X