World’s Hardest Easy Geometry Problem

Why easy? Because it can be solved only with elementary geometry. Why hard? Because the same reason, you have to use only elementary geometry. Hardest? You decide. Please send in solutions. See also these funny math problems.


Elementary Geometry

Here is everything you need to know to solve the above problems.

Lines and Angles: When two lines intersect, opposite angles are equal and the sum of adjacent angles is 180 degrees. When two parallel lines are intersected by a third line, the corresponding angles of the two intersections are equal.

Triangles: The sum of the interior angles of a triangle is 180 degrees. An isosceles triangle has two equal sides and the two angles opposite those sides are equal. An equilateral triangle has all sides equal and all angles equal. A right triangle has one angle equal to 90 degrees. Two triangles are called similar if they have the same angles (same shape). Two triangles are called congruent if they have the same angles and the same sides (same shape and size).

  • Side-Angle-Side (SAS): Two triangles are congruent if a pair of corresponding sides and the included angle are equal.
  • Side-Side-Side (SSS): Two triangles are congruent if their corresponding sides are equal.
  • Angle-Side-Angle (ASA): Two triangles are congruent if a pair of corresponding angles and the included side are equal.
  • Angle-Angle (AA): Two triangles are similar if a pair of corresponding angles are equal.

via. @ haha.nu.


25 thoughts on “World’s Hardest Easy Geometry Problem

  1. Amigo

    x=65º

    70+60=130 180-130=50 //
    70+60+20=150 180-150=30 //
    70+10+60=140 180-140=40 //

    Now you have the following ecuations
    +20 +20 +x +z =180
    +50 +x +y =180
    +30 +t +x =180
    +20 +z +t =180

    1 0 1 0 140
    1 1 0 0 130
    1 0 0 1 150
    0 0 1 1 160

    t=85 z=75 y=65 x=65

  2. George

    Actually, those problems were surprisingly easy to figure out…and it’s been a long time since I’ve done any geometry. Thanks for posting.

  3. feles

    let the central point be F
    ***
    AFB and DFE=50
    AFD and BFE=130
    ACB=20

    DCE and FBE are equal
    triangles DCE and FBE are similar, hence, DEC=130 and CDE=30
    DEA=x=20

    ***

    AFB and DFE=70
    AFD and BFE=110
    ACB=20

    DCE and DAE are equal
    triangles DCE and DAF are similar, hence, DEC=110 and CDE=50
    DEA=x=30

  4. feles

    Extratoto, I assume hahanu means that if the diagram IS drawn to scale, then x doesn’t look like 65 ;)

  5. Mulder

    Looks pretty easy – after a few simple observations, there remain 6 unknown angles with 6 equations (derived from “sum of all angles in a triangle is 180 on the plane” and “sum of all angles over a line is 180″) containing them:

    CDE, DEC, DEF (aka x), FDE, AFD (= BFE), FEB

    (where F denotes the intersection of AE with BD and CDE denotes the angle on D in the triangle formed by the point C, D and E)

    The rest is pure Gaussian elimination. Or am I missing something?

    (The “simple observations” being that ACB = 20 and AFB = DFE = 50.)

  6. hahanu Post author

    Extratoto… visually angles are drawn with their measures… and x does look smaller than ABD (which is 60)

  7. George

    For the record. The answer can’t be achieved simply by adding and subtracting the angles until straight lines equal 180* and the sum of angles within each triangle equal 180*. You have to draw a few more lines inside of the figure to get the actual answer.

    What happens if you don’t add these extra lines? Then any of several answers exist. Check and see!
    x could be various degrees and still all of the “other” calculations — based on x, or meant to check x’s value — will conform to that number.

  8. Jeff

    Ok so whats the answer? is there a page with the answers and working out on it somehwere?

    I got no idea I finished school 25 years ago. I know about the fact that all angles of a triangle equal 180 degrees but I got no idea how to get to solving X.

    Email me if you know and how its done please. zoomzone@hotmail.com

    Happy New Year.

  9. md2perpe

    By going outside of elementary geometry, i found that…
    in problem 1, x = 50 degrees,
    in problem 2, x = 30 degrees,
    i.e. the opposite to Renato.

  10. md2perpe

    Sorry, I gave the wrong numbers… That was the angle between AB and DE. The correct numbers are…
    in problem 1, x = 20 degrees,
    in problem 2, x = 30 degrees.

  11. johnny

    both fig 1 and 2 are 20 degrees because the x is as large as the top corner so if theirs 180 degrees in a triangle the answer would be 20 for both fig 1 and 2. im in 7th grade advanced algebra and i know that. its not that hard

  12. Myriam

    As for the first figure, I’d say that x, y, z and t are depending only one from the other, but the system with the 4 equations hasn’t a definite solution, all you can say is that if you give any value to x, you’ll have y = 140-x, z=x+10 and t=160-x (with y=BDE, z=EDC and t=DEC)

    Don’t know if it’s really correct, but it does sound good to me ;)

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