Sunday, May 18, 2008

Geometry Problem 1



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

Post your solutions or ideas in the comments.

33 comments:

  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.

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

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

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  4. Good problem. Here is a solution:
    *Triangle BAC and ADC are similar in this order.
    *AD is median of BC
    *So by similarity and Appolonius' theorem we get AC=sq.rt2 times DC
    and AB=sq.rt2 times AD
    *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

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

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  6. In bad english. Sorry.

    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.

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  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

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  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."

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  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

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  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.

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  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

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

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  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!

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  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.
    Which is a contradiction.

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

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

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  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

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

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  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

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  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.

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

    http://www.youtube.com/watch?v=MEv_-R0mIKc

    Greetings.

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  19. 1st problem solution

    apply sine rule to triangle ABD


    AD/sin(135-x)=BD/sin(x) then


    BD=ADxSIN(X)/SIN(135-X) PUTTING IT ASIDE


    APPLY SINE RULE TO TRIANGLE BDC

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

    BD=ADxSIN(45-X)/SIN(X) EQUATING 1&2 WE GET


    ADxSIN(X)/SIN(135-X)=DCxSIN(45-X)/SINX

    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

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

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  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

    ReplyDelete
    Replies
    1. corect link http://ogeometrie-cip.blogspot.ro/2012/08/problema-1.html

      Delete
  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
    now AD/sin(angle ABD)= BD/sin x
    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
    BD/AD=BD/DC [as AD=DC]
    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

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  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

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  24. Let AD=DC=a
    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

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  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

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  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

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  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

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  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

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  29. https://www.facebook.com/photo.php?fbid=10206189617728061&set=a.10205987640598759.1073741831.1492805539&type=3&theater

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  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.

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  31. https://www.youtube.com/watch?v=I1wHTSSUW-w

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