Saturday, February 8, 2014

Geometry Problem 976: Isosceles Right Triangle, 45 Degrees, Incenter, Angle Bisector, Hypotenuse

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 976.

Online Geometry Problem 976: Isosceles Right Triangle, 45 Degrees, Incenter, Angle Bisector, Hypotenuse

6 comments:

  1. Let D be a point on BC such that AD bisects ∠A.
    Then AID is straight line.

    ∠BID=∠BAI+∠ABI=67.5°
    ∠BDI=90°−∠BAD=67.5°
    Thus, BI=BD.

    Hence, by Problem 975,
    AC=AB+BD=AB+BI.

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  2. http://imageshack.com/a/img716/8509/6hr3.png

    On BC locate E such that BI=BE
    Triangle IBE is isosceles with ∠ (IED)=22.5 degrees= ∠ (ICD)
    Triangles IDC and IDE are congruent…. ( Case SAS) => DC=DE
    And CA= 2 x CD=CE => CA=CB+BE= AB+BI

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  3. The angle bisector BI divides isoceles right Tr. ABC. into 2 congruent isoceles right Tr. s.

    So if the in radius be r,
    AB - r = AC / 2 and BI + r = AC/2. Adding these 2 equations the result follows

    Sumith Peiris
    Moratuwa
    Sri Lanka

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  4. Problem 976
    Let point D in the extension of AB to B such that AD=AC.Then triangle ADI and ACI are congruent (AD=AC,AI=AI,<DAI=<CAI=22.5 ).So <BDI=<AC[=22.5 and <ABI=<BDI+<BID or
    45=22.5+<BID or <BID=22.5 therefore BI=BD and AC=AD=AB+BD=AC+BI.
    APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE



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  5. (More of an algebric solution)
    Let AB=BC=1 => AC = √2 and BI extended meet AC at P
    Since m(BPC) = 90, triangle BPC is isosceles right angled
    => BP = CP = 1/√2
    Semi-perimeter of ABC , S=(2+√2)/2
    S.r=.5*AB.BC => r=IP=1/(2+√2)
    => BI = BP-IP = 1/√2 - 1/(2+√2)
    => BI = 2/2+2√2
    => BI = 1/1+√2 = √2-1
    => BI = AC-AB
    => AC = AB+BI

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  6. 2r=a+b-c
    sqrt(2)BI=AB + BC - Ac =2AB-sqrt(2)AB
    so sqrt(2) (BI+AB)=2AB=sqrt(2)AC
    so BI+AB=AC

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