Friday, May 21, 2010

Problem 459: Right triangle, Squares, Distance, Measurement

Geometry Problem
Click the figure below to see the complete problem 459 about Right triangle, Squares, Distance, Measurement.

Problem 459: Right triangle, Squares, Distance, Measurement.
See also:
Complete Problem 459

Level: High School, SAT Prep, College geometry


  1. cos(GAD)= cos(GAC+CAD)=cos(GAC+45)=cos(GAC)cos(45)-sin(GAC)sin(45). But cos(GAC)=c/b & sin(GAC)=a/b hence cos(GAD)=(c-a)/(b√2). From Tr. GAD using the co-sine rule, we’ve: x^2 = GA^2 +AD^2 -2GA*AD*cos(GAD) = (c+a)^2+2b^2-2*(c+a)( b√2)(c-a)/(b√2)= (c+a)^2 + 2(c^2+a^2) -2*(c+a)(c-a)= c^2 + 2ac + a^2 + 4a^2 = 5a^2 + c^2 + 2ac

  2. Let m projection of AB over AC.
    Let n projection of BC over AC.
    Let h altitude over hypotenuse AC.
    Then (T.altitude) h = sqrt (mn)
    Then (T.legs) m = c^2/sqrt(a^2+c^2)
    n = a^2/sqrt(a^2+c^2)
    Let H point of intersection of parallel to AC from G and extension of side DC from C. Then GHD is right triangle on H.
    We've the next relations:
    CD = AC = m + n = sqrt(a^2 +c^2) (Pt. Tri.ABC)
    CH = h + n
    GH = h - n
    HD = CH + CD = h + 2n + m
    x^2=GH^2 + HD^2 = (h-n)^2 + (h + 2n + m)^2 =
    = ... = 6mn + 2*sqrt(mn)*(m+n) + 5n^2 + m^2 = ... =
    = 5a^2 + 2ac + c^2


  3. Let H point on line GF, and HF=GF. Then CG=CF and angle GCH=90.

    On triangle CDG and triangle CAH, CD=CA and CG=CH.
    angle DCG=90+angle DCH, angle ACH=90+angle DCH.

    So, triangle CDG==triangle CAH ---> then, AH=DG=x

    On triangle AGH, angle G=90, AG=a+c, GH=2a, AH=x.

    Therefore, x^2=(a+c)^2+(2a)^2 ---> x^2 = 5a^2+c^2+2ac

  4. @ third comment by Anonymous:
    CG will never be equal to CF as CG is the diagonal of the square and CF the side
    Pls correct this comment

  5. By boxing the largest square ACDE its possible to make a new right triangle with x as its hypotenuse that's easy to compute:

    First rotate and translate triangle ABC so that its hypotenuse lies on CD and its new vertex is H.
    Then extend the DH side so that it meets FG at I forming a new right triangle.

    DI = a + a
    GI = a + c
    x^2 = DI^2 + GI^2 = (2a)^2 (a+c)^2 = 5a^2 + 2ac + c^2

  6. Extend GF to H such that FH = a

    Then Tr.s DCG and ACH are congruent SAS and x^2 = AH^2 = (a+c)^2 + 4a^2 from which the result follows

    Sumith Peiris
    Sri Lanka

  7. Let AC=b. Extend DC to P such that C is the mid-point i.e. DC=CP=b. Connect FG to P and observe that FCP is congruent to CBA (SAS). Hence FP=c.
    Construct a line from D parallel to CF such that it meets GF extended at Q.
    Using mid-point theorem for the right tringle DPQ, it can be easily seen that DQ=2.CF=2a
    As DQG is a right triangle with DQ=2a, GQ=GF+FQ=a+c,=> x^2=(2a)^2+(a+c)^2
    The result follows.