Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.
This entry contributed by Sumith Peiris, Moratuwa, Sri Lanka.
Click on the figure below to view more details of problem 1261.
Friday, September 16, 2016
Geometry Problem 1261: Cyclic Quadrilateral, Circle, Diameter, Midpoint, Measurement
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https://goo.gl/photos/tDgKutmrL4Bn9Lur8
ReplyDeleteDraw points O, E, F, G per sketch
Note that OE ⊥AB , AF⊥ EF and triangles EBC, AFE are 45-45- 90 triangles
We also have AF/FC= CD/AD= 1/3 => EC//AE
And BE/BA=BC/BE= 1/2
Dear Peter - How is AF/FC = CD/AD?
DeleteSince AB=2BC=> BC=BE=AE ( see sketch)
DeleteTriangles AFE, EGB and BGC are 45-45-90 triangles and are congruent ( case ASA)
so FA=FE=EG=GC => AF/FC= 1/3
Palai Renato
ReplyDeleteWe put AB=1, BC = 2 AC= √5,
We put CD=x we have x+(3x)2=√5, and therefore x= 1/√2
<BAC =45 (regle of sum of tangent), CD=DE and CE=1
CE=CB and C is midpoint of BE
Reply
How is < BAC = 45?
ReplyDeleteABCD is cyclic, with AC being diameter. Hence, Angle ABC and Angle ADC are right angles, by Thales' theorem.
ReplyDeleteBy pythagoras’ theorem, it’s evident that BC = AC/sqrt(5), AB = 2AC/sqrt(5), CD = AC/sqrt(10), AD = 3AC/sqrt(10). We need the sine and cosines of angle BAC, angle CAD.
If we let O be the circumcenter of ABCD, then angle BOC = 2 angle BAC, angle COD = 2 angle CAD. This is obvious high school geometry.
From here, we use complex numbers.
Let the circle on ABCD be the unit circle and c = 1, a = -1.
With a bit of trig (especially double angle formula), b = cis(angle BOC) = cis(2 angle BAC) = (3/5) + i(4/5). Similarly, d = cis(-angle COD) = (4/5) – i(3/5).
Now let’s make a point F such that C is the midpoint of B and F. This indicates (f + b)/2 = c, which rearranges to f = 2c – b. We compute f = (4/5) – i(4/5).
We now assert that A, D and F are collinear. This is proved by checking that (a – f)/(a – d) is a real number (justification is De Moivre theorem; (a - d) multiplied by a real number would make (a-f) iff a-f and a-d make an angle of 0 or 180 degrees with each other).
(a – f)/(a – d) = (12/5 – i(4/5))/((-9/5) – i(3/5)) = (4/5) / (-3/5) = -4/3 which is definitely a real number. So we conclude that F is on line AD and BC (as C is midpoint of BF).
But we know that E is the intersection of AD and BC. This means E = F, and the proof is complete.
Problem 1261
ReplyDeleteIs <BAC+<CAD=45, so <CED=45 then CD=DE. Let the points K, L on the side AD such that
AK=KL=LD=AD/3=DC=DE. So AL=LE and BL is perpendicular at AE so BL//CD. Therefore
BC=CE (LD=DE).
APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE
Palai Renato,
ReplyDeleteReply to Sumith Peiris to request to motivate why angle.. is 45°.
If we calla <BAC= and <CAD= , <BAD= + , bat
tang ( + )=(tang + tang )/(1 - tang tang ) and introducing the values of and , we have: (2+3)//1-2x3)= -1, tang 1 = 45°
Reply
Thank u. Good pure geometry proof
ReplyDeleteGood pure geometry solution
ReplyDeleteLet CD = 1
ReplyDelete=> AD = 3
=> AC^2 = CD^2 + AD^2 = Sqrt(10)
Similarly solving for AB & AC in the right triangle ABC,
we get BC = sqrt(2) --------(1)
and AB = 2*Sqrt(2)
Let CE = x and DE = y
As CE^2 = DE^2 + CD^2
=> 1+y^2 = x^2 ----------- (2)
As ED.EA = EC.EB (From Secant-Secant Rule)
=> y.(3+y) = x.(x+Sqrt(2))
=> y^2+3y = x^2+Sqrt(2).x
=> 3y = (x^2-y^2) + Sqrt(2).x
=> 3y = 1 + Sqrt(2).x (from (1) )
=> y = [1+Sqrt(2).x]/3 ----------(3)
Substituting the value of y in Eq(2)
=> 9 + (1+2x+2Sqrt(2)x) = 9x^2
=> 7x^2 - 2Sqrt(2)x -10 = 0
Therefore x = (2Sqrt(2) + Sqrt(8+288))/14 = 14.Sqrt(2)/14 = Sqrt(2) = BC (from (1) )
Hence C is midpoint of BE
Let BC = x
ReplyDeletethen AB = 2x
AC^2 = x^2 + (2x)^2
AC^2 = x^2 + 4x^2
AC = sqrt(5)x
Let CD = y
then AD = 3y
AC^2 = (3y)^2 + y^2
AC^2 = 9y^2 + y^2
AC = sqrt(10)y
sqrt(5)x = sqrt(10)y
y = sqrt(5)x/sqrt(10)
= sqrt(5)x*sqrt(10)/sqrt(10)*sqrt(10)
= sqrt(50)x/10
= 5*sqrt(2)x/10
= sqrt(2)x/2
Let CE = b
then y^2 + DE^2 = b^2
DE = sqrt(b^2 - y^2)
In tri CDE and tri ABE
<ABC = <CDE (diameter AC)
<AEB = <DEC (common angle)
<BAC = <DCE (remaining angles of a tri)
therefore triangle ABE and triangle CDE are equiangular triangles
BE/DE = AB/CD
(x + b)/sqrt(b^2 - y^2) = 2x/x*sqrt(2)/2 = sqrt(8)
x+b = sqrt(8)*sqrt{b^2 - (x*sqrt(2)/2)^2}
(x+b)^2 = 8(b^2 - x^2/2)
x^2 + 2xb + b^2 -8b^2 +4x^2 = 0
5x^2 + 2xb - 7b^2 = 0
5x^2 + 7xb -5xb - 7b^2 = 0
x(5x+7b) -b(5x+7b) = 0
(x-b)(5x+7b) = 0
x-b = 0
x = b
CE = b = x
CE = BC = x
therefore C is the midpoint of BE
Let CD = x and BC = y
ReplyDeletethen AD = 3x and AB = 2y
(2y)^2 + y^2 = (3x)^2 + x^2
5y^2 = 10x^2
y = sqrt(2)x
Let CE = t and DE = p
then x^2 + p^2 = t^2
t = sqrt(x^2 + p^2)
(2y)^2 + (y + t)^2 = (3x + p)^2
4y^2 + y^2 + 2yt + t^2 = 9x^2 + 6xp + p^2
4(sqrt(2)x)^2 + (sqrt(2)x)^2 + 2(sqrt(2)x)(sqrt(x^2 + p^2)) + (sqrt(x^2 + p^2))^2 = 9x^2 + 6xp + p^2
8x^2 + 2x^2 + 2x(sqrt(2))(sqrt(x^2 + p^2)) + x^2 + p^2 = 9x^2 + 6xp + p^2
2x(sqrt(2))(sqrt(x^2 + p^2)) = 6xp - 2x^2
(sqrt(2x^2 + 2p^2)) = 2x(3p - x)/2x
2x^2 + 2p^2 = 9p^2 - 6xp + x^2
x^2 + 6xp - 7p^2 = 0
x^2 + 7xp - xp - 7p^2 = 0
x(x + 7p) - p(x + 7p) = 0
(x - p)(x + 7p) = 0
x - p = 0
x = p
CD = DE
<DCE = <DEC = 45 (Isosceles tri theorem)
<DEC = 45 = <EAB (Sum of angles in a tri)
BA = BE
But BA = 2BC
thus BE = 2BC
BE = BC + CE
2BC = BC + CE
BC = CE
therefore C is the midpoint of BE
Draw EP ꓕ AC, from similarity CP = a, PE = 2a, AC = 5a
ReplyDeleteFrom ∆ABC => a = x/√5, from ∆PEC => EC = x
From: Sumith Peiris. March 1, 2019 at 3:07 AM
ReplyDeleteTo c.t.e.o
Problem 1261
Mark X on AB such that BX = BC
Mark Y on AD such that DC = DY
Use Pythagoras and show CX = AY and AX = CY that is AXCD is a parallelogram.
X mid point of AB so C midpoint of BE
Ok thanks
ReplyDelete<ABC=<ADC=90
ReplyDeletetan<BAC=BC/AB=1/2
tan<CAD=CD/AD=1/3
tan<BAD=(1/2+1/3)/(1-(1/2)*(1/3))=1
<BAD=45
<BCD=135
<DCE=45
CD=DE
Let CD=DE=a
CE=a*(sqrt2)
AD=3a
AC=a*sqrt(10)
Let BC=b
b^2+(2b)^2=AC^2=10a^2
5b^2=10a^2
b^2=2a^2
b=a*(sqrt2)
So, b=BC=a*(sqrt2)=CE
C is mid-pt of BE