## Friday, September 16, 2016

### Geometry Problem 1261: Cyclic Quadrilateral, Circle, Diameter, Midpoint, Measurement

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. 1. https://goo.gl/photos/tDgKutmrL4Bn9Lur8

Draw 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

1. Dear Peter - How is AF/FC = CD/AD?

2. Since AB=2BC=> BC=BE=AE ( see sketch)
Triangles AFE, EGB and BGC are 45-45-90 triangles and are congruent ( case ASA)
so FA=FE=EG=GC => AF/FC= 1/3

2. Palai Renato
We 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

3. How is < BAC = 45?

4. ABCD is cyclic, with AC being diameter. Hence, Angle ABC and Angle ADC are right angles, by Thales' theorem.

By 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.

5. Problem 1261
Is <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

6. Palai Renato,
Reply to Sumith Peiris to request to motivate why angle.. is 45°.
tang ( + )=(tang + tang )/(1 - tang tang ) and introducing the values of and , we have: (2+3)//1-2x3)= -1, tang 1 = 45°

7. Thank u. Good pure geometry proof

8. Good pure geometry solution

9. Let CD = 1
=> 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

10. Let BC = x
then AB = 2x

AC^2 = x^2 + (2x)^2
AC^2 = x^2 + 4x^2
AC = sqrt(5)x

Let CD = y
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

11. Let CD = x and BC = y
then 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

12. Draw EP ꓕ AC, from similarity CP = a, PE = 2a, AC = 5a
From ∆ABC => a = x/√5, from ∆PEC => EC = x

13. From: Sumith Peiris. March 1, 2019 at 3:07 AM
To 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

14. 