See complete Problem 46
Triangle, Angles, Midpoint, Congruence. Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
Monday, May 19, 2008
Elearn Geometry Problem 46
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Solucion:
ReplyDeletehttp://img8.imageshack.us/img8/9236/rsolmarzo7.jpg
In Tr. DBC, DC/DB = sin(3x+30)/sin(30) = 2sin(3x+30) While Tr. ADC gives,DC/AD =sin(2x)/sin(x)
ReplyDeleteNow AD=DB gives, 2sin(3x+30)=sin(2x)/sin(x) or
2sin(3x+30)=2sin(x)cos(x)/sin(x) or
sin(3x+30)=sin(90-x) or 3x+30=90-x or x=15
Alternatively, sin(3x+30)=sin(90-x) may mean
sin(180 -(3x+30))=sin(90-x) giving us,
180 -(3x+30) = 90- x giving us x=30
Svvv is very correct in saying that 15 deg. and 30 deg. are both acceptable solutions to this problem.
Ajit: ajitathle@gmail.com
Thanks for the comments.
ReplyDeleteNew condition: angle B is an obtuse angle.
Dear Antonio
ReplyDeleteThis problem is proving very difficult to crack thro Euclidean geometry alone.
I tried to drop altitudes from B to DC and AC and work through the cyclic quad
Secondly tried to find point M on DC such that D is the centre of ABM Tr.
Thirdly found that if u find N on DC such that Tr. ANC is isoceles I found that < DBN is 30
But I cannot clinch it
Are any of my approaches right?
Dear Sumith,
DeleteSome ideas: equilateral triangle, isosceles triangle, congruence ....
Let BE = h be an altitude of triangle ABC and let CD = d
DeleteSo BD = DA = DE = CE = c/2
CD is a median of triangle ABC, so
d^2 = a^2/2 +b^2/2 -c^2/4 ....(1)
Since triangles BDC and ADC are equal in area, easily
1/2 da/2 = 1/2 bh/2 hence d = bh/a ....(2)
Using Pythagoras h^2 = c^2 - (b-c/2)^2 = a^2 - (c/2)^2....(3)
Eliminating d and h between these 3 equations,
a^2/2 + b^2/2 - c^2/4 = (b^2/a^2)(a^2 - c^2/4).
Simplifying
a^/2 - c^2/4 = b^2/2 - b^2c^2/(4a^2) which gives,
2(a^2 - b^2) = (c^2/a^2)(a^2 -b^2)
Since a NOT = b, a^2 = (c/2)c
So BC^2 = BD. BA and BC is therefore a tangent to circle ADC.
Hence 2x = 30 and x = 15
Sumith Peiris
Moratuwa
Sri Lanka
Draw the altitude BE in triangle ABC. Draw DE. AD=DE=DB. => ang(DEA)=ang(DAE)=2x In triangle DEC is ang(EDC)=x ; DE=EB. Now mirror D in BC that gives point F and triangle DFC is equilateral. DB=BF. Hence triangle(DFB) equals triangle(DCE) (SSS: DB=BF=DE=EC and DF=DC) => ang(FDB)=x; ang(BDC)=3x
ReplyDeleteang(FDC)=4x => 4x=60 => x=15.
This comment has been removed by the author.
ReplyDeleteThis comment has been removed by the author.
DeleteSolution 2
ReplyDeleteLet the perpendicular bisector of CD meet CD at E, AC at F & let the center of Tr. CBD, (obtuse at B) be G with G obviously on EF situated outside Tr. CBD
Since < DGB = 60, Tr. BGD is equilateral. Hence DF = DG which is impossible
Thus F and G coincide < BFD = 60 and so 2x = 30 and x = 15
Sumith Peiris
Moratuwa
Sri Lanka
Solution 3
ReplyDeleteDrop a perpendicular BP from B to AC, P on AC. Draw equilateral CDQ on DC.
Now AD = DP = PC = DB = BQ and so Tr.s CDP & BDQ are congruent SSS
So < BDQ = 60-3x = < PCD = x from which,
x = 15
Sumith Peiris
Moratuwa
Sri Lanka
If B is obtuse x = 15
DeleteIf B is acute x = 30
If B is right x = 20
Hello Sumith,
DeleteMy finding is that B cannot be a right angle. You can check my post below for details.
Euclidean regards.
Greg
It is interesting to note that if the condition “angle B is an obtuse angle” were to be lifted, which was the case in the original problem statement, the 2 solutions x=15° or x=30°, that Ajit found at the beginning through trigonometry, can be determined in various “Euclidean” ways.
ReplyDelete1/ In Sumith Peiris first proof : the statement “since a NOT b” is required to get rid of the 0=0 trivial solution to the equation, but one could very well look at what happens when a=b and find that in that case x=30°, which is a valid solution.
2/ In Piet van Kampen’s proof : after finding that DFC is equilateral, “ang(FDB)=x; ang(BDC)=3x ang(FDC)=4x” because B is obtuse, but if B were acute, then F would be, on BC, on the other side of DB and the series of equations would be “ang(FDB)=x; ang(BDC)=3x ang(FDC)=2x” which gives x=30°, which is a valid solution.
[In the case where B would be allowed to be a right angle, Piet’s proof would not work because D, B and F would be aligned and triangles DFB and DCE would not be equal triangles anymore, so that the above equations would not stand. And in that case, when B is a right angle, one may wonder whether x=20°could be a third valid solution.]
3/ Lastly, a different way of looking at the problem could be the following (see animated graph here).
Allow ang(BCD) to take any value, the remaining of the problem’s construction being unchanged, and define E as the foot of the altitude from B : then AD=BD=DE (ABE right triangle in E) and DE=EC because ang(CDE)=pi-x-(pi-2x)=x.
Draw circle C1 of diameter AB, and circle C2 of center E and radius EC=ED, and allow E to “move” from A to B on arc(AB), keeping C on AE such that EC=ED=AD : the problem is equivalent to finding the positions of E that make ang(BCD) to be equal to 30°.
Let G be the intersection of C1 and C2 on arc(AB) : DEG is equilateral (because DE=DG in C1 and ED=EG in C2) therefore ang(DCG) = ang(DEG)/2 = 30°. So as E “moves” on arc(AB) from A to B, the equilateral triangle DEG rotates around D within C1 and G “moves” on arc(AB) ahead of E towards B.
For ang(DCB) to be equal to 30°, as ang(DCG)=30°, B, C and G must be aligned (even allowing G and B to coincide, as the case may be).
If G and B do not coincide, since BDG is isosceles in D and CEG is isosceles in E, Ang(DBC)=ang(DBG)=ang(BGD) and ang(ECB)=ang(ECG)=ang(CGE) therefore ang(DBC)+ang(ECB) = ang(BGD)+ang(CGE) = pi -ang(DGE)=2*pi/3 therefore 2x=pi/3 and x=30° (first solution – first stop of the animation).
If E is allowed to move further on arc(AB), at some point G becomes equal to B and ang(DCB)=ang(DCG) = 30° which is also a valid configuration. In that case BDE is equilateral and ang(BDE)=4x=60° therefore x=15° (second solution – second stop of the animation).
So there are 2 valid solutions to the original problem, x=30° or x=15°.
And the answer to the question about x=20° being a valid solution or not (see 2/ above) is : it cannot be because when B, G and C are aligned : ang(ABC)=ang(ABG)=ang(ADG)/2. By our construction, ang(ADG) can only be equal to pi when G=B which is when x=15° and ang(ABC)=105°. And this is confirmed by the attached graph (simple trigonometry would also confirm it easily).
If x=20 then it is possible to draw a triangle ABC such as Ang(A)=2x, Ang(ACD)=x, Ang(DCB)=30.
DeleteBut, in that case, DA will not be equal to DB…
See it in this drawing
Draw the circle C1, with incenter D and radius DA=r1
ReplyDeleteDraw the circle C2, passing through D, B and C, with incenter 0 and radius DO=r2 : OD=OB=OC
Let suppose O is not on AC, neither on C1
ang(BCD) = 30° intercepts arc DB => ang(DOB) = 60°
ang(DOB) = 60° , OD=OB => ΔDOB is equilateral
Therefore DO=DB=r1 => O is on C1
ΔDOB is equilateral => DO=DB and DB=DA => DO=DA
O is on C1 => ang(AOB) = 90°
ang(AOB) = 90° , ang(BOD) = 60° => ang(DOA) = 30°
Define E, the intersection of C1 and AC. ang(AEB) = 90
DA=DB=DE therefore ΔADE is isosceles in D => ang(AED) = 2x
ang(DAE) = 2x and ang(DCA) = x => ang(CDB) = 3x
ang(EDB) = ang(DAE) + ang(DEA) =4x
ang(EDC) = ang(EDB) - ang(CDB) = 4x-3x = x
ang(EDC) = x and ang(ECD) = x therefore ΔCED is isosceles in E => ED=EC
O=E?
What we know about E : E on C1, on AC, ang(AEB) = 90 and ED=EC
What we know about O : O on C1, ang(AOB) = 90 and OD=OC=OB
Define circle C3, with incenter C and radius C0=CE=r1
It means that E and O are on C1 and C3
Let's examine the various situations with respect to the intersection(s) of C1 and C3:
Option 1 : There are only two points, intersection of the two circles C1 and C3 with the same radius r1
Option 1a : E≠O :
let suppose E and O are on these two different intersections (Cf. graph)
OD=OC=ED=EC => ODEC is a diamond => DC is the angle bisector of ang(EDO) and ang(ECO)
ang(EDC)=x => ang(CDO)=x
ang(CDB)=3x, ang(CDO)=x => ang(ODB)=2x
Since ΔDOB is equilateral => 2x=60 => x=30
Option 1b : E=O
If O=E, then ang(DEA) = ang(DOA), 2x=30 => x=15°
Option 2 : There are only one point, intersection of the two circles C1 and C3
Then E=O Cf. Option 1b
Option 3 : There is no point, intersection of the two circles C1 and C3
Since EC=ED => E belongs to C1 and C3 : this option is impossible
Conclusion : 2 solutions, x=15° and x=30°. See graph
This is a link to the drawings