## Sunday, May 18, 2008

### Geometry Problem 1

See complete Problem 1
Triangle, Angles, Midpoint, Congruence. Level: High School, SAT Prep, College geometry.

1. Problem 1
In french, sorry...
On note H le pied de la hauteur issue de B.
On pose BH = h, AD = AC = 1.
En utilisant la trigonométrie dans les triangles ABH et BCH, rectangles en H, on obtient :
tan x = h/(1 - h)
et tan(45 -x) = h/(1 + h)
En utilisant tan(a - b) = ....
on arrive à l'équation :
2h² + 2h-1 = 0
qui a 2 solutions.
La solution positive est
(SQRT(3)-1)/2
Ce qui donne tan x = SQRT(3)/3
D'où x = 30°

ChrG.

2. consider similar triangle BDC and ABC,
DC/BC=BC/AC => DC/BC=1/√2=sin{x}/sin{45°} => x=30°

3. Could you please tell on how did you get DC/BC as 1/sqrt(2)

4. Good problem. Here is a solution:
*Triangle BAC and ADC are similar in this order.
*So by similarity and Appolonius' theorem we get AC=sq.rt2 times DC
*AM be perpendicular on BC. (B-M-D-C)
*Triangle AMD is 45-45-90.
*So AD = sq.rt2 times AM.
*So AB = sq.rt2 times sq.rt2 times AM. i.e. AB = 2 times AM.
*Now AMB is right angle triangle,
where hypotenuse (AB) is two times one of the sides (AM). Hence is has to be 30-60-90 triangle.
*SO FINALLY ANGLE x = 30 degrees

5. Small correction
Triangles BDC and ABC are similar in that order.

Let O the center of a circunference that contains points A,B and C. Extend segment BD to intersect circunference at point F. So ...

1. DO is perpendicular to AC. (D is midpoint)
2. Arc BC = Arc CF
3. CO is perpendicular to BF, intersecting at point E.
4. Triangle CED is retangle and isosceles.
5. Triangle OED is retangle and isosceles.
6. Triangle COD is retangle and isosceles.
7. OE = EC
8. Angle AOB = Angle EBC
9. sen x = OE/OB = 1/2

Then, x = 30 degrees.

Marcio, from Rio de Janeiro.

7. Apply sine rule in Tr. BDC to get DC/BC=sinx/sin45 =V2sinx where V2=square root of 2. Likewise in Tr. ABC, BC/AC=V2sinx and 2DC = AC; hence 2V2sinxBC = BC/(V2sinx) or (sinx)^2 = 1/4.
Therefore, sinx=1/2 ignoring the negative sign since angles in a triangle can only be +ve and thus x=30 deg.
How may one prove this without using trigonometry?
Ajit: ajitathle@gmail.com

8. Dear Joe,
It's possible to solve this problem without using trigonometry.
The key is auxiliary construction:

Geometry problem solving is one of the most challenging skills for students to learn. When a problem requires auxiliary construction, the difficulty of the problem increases drastically, perhaps because deciding which construction to make is an ill-structured problem. By “construction,” we mean adding geometric figures (points, lines, planes) to a problem figure that wasn’t mentioned as "given."

9. How about this for a plane geometry solution? From C drop a perpendicular to AB extended meeting it in N. Let CN = p. Angle NBC=45 from the given data. Hence angle BCN=45 and therefore BN=p and by Pythagoras, BC^2 = p^2+p^2=2p^2
Now triangles BDC & ABC are similar with angle BCD as the common angle & angle DBC=x=angle CAB. Hence AC/BC=BC/DC or BC^2=AC*DC=2DC^2 since D is the midpt. of AC. But as shown above BC^2=2p^2. So 2DC^2=2p^2 or DC =p. In other words, AC = 2p. Now in the rt. angled triangle ACN we've CN=p and AC=2p; hence ACN is 30-60-90 with angle NAC =30 deg. or x = 30 deg.
Ajit
ajitathle@gmail.com

10. Los triángulos ABC y BDC son semejantes por tener dos ángulos iguales. Sea AB=c, AC=b,BC=a Entonces c/BD=b/a=a/(b/2) De aqui se ve que b^2=2a^2 (b^2 es b al cuadrado) y BD=ac/b Al elevar al cuadrado esta igualdad y usando la otra igualdad se obtiene que BD^2 = (c^2)/2 Si se traza la altura BX se forma un triángulo rectángulo isósceles de catetos h Por Pitagoras se tiene que BD^2=2h^2 y comparando con la ultima igualdad se ve que c^2 = 4(h^2) y al sacar raíz cuadrada vemos que c = 2h En el triángulo rectángulo ABX la hipotenusa AB es el doble del cateto BX Luego, el ángulo opuesto a BX = h es 30º. Esto tiene demostracion no trigonometrica.

11. sorry, my english is not god.i'm brazilian, but i think i have a better way to do it.i'll write it in portuguese:
prolongando o lado AB e formando um angulo de 90 com o lado AC.Assim teremos a mediana BD como mediana relativa a hipotenusa(sendo assim igual a mediana do triangulo ABC).ligando o angulo reto jah formado, perceberemos q este é isosceles, e formará outro isósceles EDC, logo, mexendo com os angulos(o q n é tão dificil) vc provará q este triangulo é equilatero mostrando q 2x= 60, logo = 30

12. http://ahmetelmas.files.wordpress.com/2010/05/cozumlu-ornekler.pdf

13. Note as in other solutions that triangle BDC is similar to triangle ABC, thus DC/BC=BC/AC. Let AD=DC=t, thus AC=2t.

Cross-multiplying gives BC^2=2t^2, and BC=t*sqr(2).

Now drop an altitude from B in triangle ABD and from D in triangle BDC, and call them y and h, respectively. Since a median divides a triangle into two triangles with equal areas, yt=h(t*sqr(2)). solve for h=(sqr(2)/2)*y.

But the altitude from B creates a right, isosceles triangle, with hypotenuse BD, thus BD=sqr(2)*y. This means BD=2h, making a 30-60-90 triangle. QED using no trigonometry!

14. Here another approach: Assume x is not 30.

Let A', C' be such that <DA'B=30 and <DBC'=30.
(Triangle A'BC' is obtained by joining two equilateral triangles in opossite sides of a square.)
Now, it is easy to show that BD is also median of BA'C'.

Case 1: If A' is between A and D, by Euclid I.16, angle BAD is less than angle BA'D=30
but also angle DBC is greater than angle DBC'=30
As a consequence both angles can't be equal.

Case 2: If A is between A' and D the proof is analog.

Therefore A=A', C=C' and the result is clear.
:)

15. Let O be the circumcentre of triangle ABC.
Produce BD to meet OC at E and the circumcircle at F.
Triangle FOC is isoceles with base angle 2x.
The semivertical angle of triangle is x.
X = 30

1. How do you presume point B lies on Circle O?

16. since BDC and BAC are similar
DC/BC=BC/AC
let DC=a
then BC=sqrt2 * a
Draw a half circle having it's Centre D and Diameter is AC
Extend AB to meet the half Circle at E
ABC=135 Deg
so, EBC=45 Deg
Since E=90Deg
ECB=45Dg
so BE=EC=a (As CB=sqrt2 * a)
Now Joint ED
ED=a (ED is another radius of half Circle)
so, EDC is eqlateral
So, EDC=60 Deg
as EDC=2*EAC
EAC=30 Deg

17. The easiest and most succint solution:

Triangle ABC & triangle BDC are similar.
Therefore, BC/CD=AC/BC=AB/BD.
Taking BC/CD=AC/BC, we get, BC.BC=2CD. CD
So, BC=sqrt (2). CD

Taking BC/CD=AB/BD, we get, AB=sqrt (2). BD
Let, AB= c
So, BD=c/sqrt (2)

Dropping an altitude BE in triangle ABD, we get a right isosceles triangle whole hypotenuse is BD and BD=c/sqrt (2)

So the legs: BE=ED=c/2

Now we see that the triange ABE is a right triange with sides AB=c, BE=c/2 and AE=c. sqrt (3)/2

This clearly implies
ABE to be a 30-60-90 triangle.

Therefore, Angle BAE= x= 30 degree.

18. Two solutions, the first bye me, and the second by my friend.

Greetings.

19. 1st problem solution

apply sine rule to triangle ABD

APPLY SINE RULE TO TRIANGLE BDC

DC/SIN(X)=BD/SIN(45-X)

SIN2(X)=2SIN(45-X)xCOS(45-X)/2

SIN2(X)=SIN(90-2X)/2=COS(2X)/2

SIN2(X)=1-SIN2(X)/2

4SIN2(X)=1

SIN(X)=1/2

:X=30

20. Video solution (Spanish version) by Eder Contreras and Cristian Baeza at Geometry problem 1.

21. Solution sent by Petre CIOBANU

Hello
another solution of #1 here
unfortunately is now written in Romanian (a language of Latin origin).

I respected the request to use the basic methods.
I write here because I could not post the message box with solutions.

I would be glad that you appreciate my solution.

Thanks and congratulations for the site,

Petre CIOBANU

22. angle BDC = 180 - angle BDA
= 180 - 45 = 135 degree
we also know that angle BDC = angle DAB + angle ABD
= x + angle ABD
ie angle ABD = (135-x)degree
or BD/AD = sin x/ sin(135-x)........... 1
for triangle BDC
DC/sin(angle DBC)=BD/sin (angle BCD)
BD/DC= sin (angle BCD)/sin(angle DBC) = sin(45-x)/sin x...............2
from 1 and 2 we get
sin x/ sin(135-x)=sin(45-x)/sin x
or (sin x)^2 = sin(135-x)sin(45-x)
or (sin x)^2 = 1/root2(cos x + sin x) . 1/root2 (cos x - sin x)
or (sin x)^2 = 1/2[ (cos x)^2- (sin x)^2]
or 2.(sin x)^2=[ (cos x)^2- (sin x)^2]
or 3.(sin x)^2=(cos x)^2
or (tan x)^2 = 1/3
or tan x =1/root3
or x=30 degree

23. Let AD = DC = t.
As seen above, [BDC] and [ABC] are similar => BC = sqrt(2) t.

Drop a perpendicular from C to AB produced cutting it at E and let BE = s.

By considering angle(EBD) as an exterior angle to [ABD], we have angle(EBC) = 45 degrees.
EBC is then an 45–45–90 triangle with EC = s and BC = sqrt(2) s
Thus, s = t

In [AEC],
EC = t, AC = 2t and angle(E) = 90 degrees => AEC is an 30–60–90 triangle.
Therefore, x = 30 degrees

Mike from Canotta

Since BDC and ABC are similar BC=V2a
now arBAD = arBDC or 1/2. a. BD. sin 45 = 1/2. BD. V2 a. sin x
sin x = 1/2 since x is acute so x = 30 degrees

25. This is a very good problem
here is the solution hope you understand it
Stratergy:In this problem we need to show that Tri BAEis 30-60-90(BE IS THE ALTITDE)for this we will use the thereom if one side of an right angeled Tri is half the hypotenuse the opp Ang is =30
PROOF
TriBDC and TriABC are similar
BC/AC=DC/BC=BD/AB-----1
considerDC/BC=BC/AC
let DC=a
then BC=sq
CONSTRUCT Tri BED(BE IS THE ALTITDE)
LET BD=y
Th EB=ED=y/sqrt2---2
Using 1
BA/BD=BC/DC
sqrt2*y/y=AB/y
Th sqrt2*y=AB-----3
Using the Stratergy
if BE*2=AB
THE PROOF IS COMPLETE
USING 1,2,3TO FILL VALUES WE GET
Y/sqrt2*2=BE
sqrt2*sqrt2*x/sqrt2=BE(2=sqrt2*sqrt2)
simplifying we get
BE=sqrt2*x
AB=sqrt2*x
AB=BE
Using Stratergy
x=30
Hence proveD

26. Take P - symmetrical of C about AB; clearly tr. BCP is right-angled and isosceles, hence <BPC=45=<ADB and BDCP is cyclic, so PD_|_AC, i.e. PC=AP, but by symmetry AP=AC and ACP is an equilateral triangle, or <BAC=30 degs. Best regards, Stan Fulger

27. Since Tr.s ABD & BDC are equal in area, BD/sqrt2 X AD = DE X BC where DE is the altitude. But since BC is tangential to Tr. ABD, BC = sqrt 2 X AD from which we deduce that DE = 1/2 BD and so x = 30

Sumith Peiris
Moratuwa
Sri Lanka

28. The symmetrical of AC about AB and BC respectively intersect at E; since <BAC+<BCA=45, we infer <AEC=90 and B is the incenter of triangle AEC, hence <AEB=45=<ADB, consequently triangles AEB and ADB are congruent (a.s.a criterion), making AE=AD. From right-angled triangle AEC we get DE=AD, triangle ADE equilateral and <EAD=60, consequently <BAD=30

30. The circumscribed circl with triangle BCD intersects the perpendicular bisector AC in E. Then AE=EC and <DAB=<DBC=<DEC=x. If AB intersects EC in K so <DAK=<DEK=x .Is BK perpendicular EC(<KBC=<KAC+<BCD=<DBC+<BCD=45=<KBE ),then BC=BE, EK=KC. But triagle AKC=triangleAKE,so AE=AC <BAC=x=<BAE and triangle AEC it is equilateral. Therefore 2x=60
Or x=30.

32. Let E, be a point on (AD) ̅ such that (BE) ̅⊥(BD) ̅.
Let F, be a point on (AD) ̅ such that (BF) ̅⊥(AD) ̅.
m∠EBD=m∠BFD=m∠BFE=90°
∆EBD,∆BFD and ∆BFE are isosceles right triangles.
BF=EF=FD=a, FA=b