Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.
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Friday, December 7, 2018
Geometry Problem 1406: Four Congruent Circles, Sum of the Squares, Measurement, Metric Relations
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If FP = d and BF = r, following on from my proof for Problem 1405,
ReplyDeleteUsing Appollonius in Tr. AFP,
a^2 + d^2 = 2.b^2 + 2.r^2 .....(1)
Using Ptolemy in CBFP,
Sqrt3.b = c + d ......(2)
Using cosine rule in Tr. CFP
c^2 + d^2 - cd = 3.r^2 .....(3)
From (2) & (3), by eliminating d & simplifying
c^2 + d^2 = 2r^2 + b^2 .....(4)
From (1) & (4) the result follows
Sumith Peiris
Moratuwa
Sri Lanka
ABF straight line.United P with F,the straight line PF cut the circle with center B in pont H.˂ AHF=90º . PB=AH , PC=PH. Δ APH AP²=AH²+HP²=PB²+PC². a²=b²+c²
ReplyDeleteErina,NJ
It is easy to see that Angle BPC= 30 Deg
ReplyDeleteNow rotate Points P and C clockwise around point B by 60 Deg.
Point C will move to A and point P moves to P'.
Triangle BP'A is congruent to triangle BPC hence Angle BP'A=30 Deg and AP'=PC=c.
also Triangle BPP' is equilateral, hence Angle BP'P=60 Deg and PP'=BP=b.
Triangle APP' is right triangle with AP=a as hypotenuse. Hence a^2=b^2+c^2.
Join P to S (S intersection D and C). PS meet C at R. Tr PBR equilateral(my solution 1405)
ReplyDeleteExtend PF to T ( T on B) Tr PTC equilateral (TPC=PTC=60)
=> TPR=PTA=90 => TARP rectangle => TPR right tr with our three sides