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Fun math to teach

  1. Sep 4, 2011 #1
    I've got to teach a class of 10th graders math tomorrow. I was thinking of showing some fun things in math like puzzles, understandable open problems in math,etc.

    Can anyone suggest me some fun stuff in math that would captivate 10th graders?
  2. jcsd
  3. Sep 4, 2011 #2
    what is the curriculum that they are supposed to master?
  4. Sep 4, 2011 #3
    Nah, I'm actually a high-schooler, and we have this Teachers' Day in our school, where we seniors teach the younger kids. So, it's just supposed to be a mock class, but I wanted the kids to take an interest in math,not just listen to a boring guy with equations, because I myself am in love with math, and want others to love it too. :D ...so what do you suggest?
  5. Sep 4, 2011 #4
    What topics did you learn when you were a sophomore?
  6. Sep 4, 2011 #5
    Well, I know sets,trigononometry,sequences & series,2-D geometry(circles,lines,ellipses,hyperbola,parabola), matrices & determinants, basic differential calculus, and tidbits on prime numbers.
  7. Sep 4, 2011 #6
    WOW! All of this in grade 10?! You must be geniuses in your school.

    May I suggest a geometry problem:


    Using only elementary geometry, determine angle x. Provide a step-by-step proof.

    Attached Files:

  8. Sep 4, 2011 #7


    Let angle CDE=y
    angle CED=z

    Let that almost central point in the triangle be denoted by O.

    Then, angle CDB=180-(20+20)=140 [Angle sum property of a triangle]

    angle CEA=180-(20+10)=150[Angle sum property of a triangle]

    angle DOE= 50[Vertically opposite angles]

    ==>y+z=160[Angle sum property of a triangle]----eq 1

    x+z=180-(20+10)=150[Angle sum property of a triangle]----eq 2

    eq 1-eq 2
    ==> y-x= 10 [eq 3]

    ==>x+y=130 [eq 4]

    eq 3 + eq 4
    ==> 2y=140

    Sub in eq 3,


    Whew!...Sorry I took a long time...had difficulty typing on
    my cramped iPod...Also,please forgive my crazy notations...

  9. Sep 4, 2011 #8
    How did you deduce that?
  10. Sep 4, 2011 #9
    Oops, I made a mistake...Give me a couple minutes,please..
  11. Sep 4, 2011 #10
    I'm sorry, but I'm unable to find x(I must seem like a total idiot). Could you please complete the proof?
  12. Sep 6, 2011 #11


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    Science Advisor
    Homework Helper

    Line CD has to be equal to line DB.

    Line DE is common to triangles CDE and BDE.

    Try the law of sines for each triangle. There will be two possible answers mathematically, but the 'correct' answer will be obvious, and will also show why you got the correct answer in spite of your error.
  13. Sep 7, 2011 #12
    Ah, thank you BobG! Can't believe I overlooked something as simple as this !

    So angle DCE=180-(80+80)=20

    so that CD=DB(Isosceles triangle property)

    then,by the law of sines,

    CD/sin(z) = DB/sin(x+30)
    ==>z-x=30----eq 1

    and from earlier, z+x=150---eq 2

    eq 2-eq1
    Last edited: Sep 7, 2011
  14. Sep 7, 2011 #13
    Actually, by the Law of sines, you would get:
    \frac{\overline{CD}}{\sin{(z)}} = \frac{\overline{DE}}{\sin{(20^\circ)}} = \frac{\overline{DB}}{\sin{(x + 30^\circ)}}
    which, by the isosceles property that you noticed, would make:
    \sin{(z)} = \sin{(x + 30^\circ)}

    Now, if the sine of two angles is the same, it means that either:
    z = x + 30^\circ
    which is what you call Eq.(1)

    z = 180^\circ - (x + 30^\circ) = 150^\circ - x
    which gives back your Eq.(2).

    So, you how do you know that we have the first case and not the second?
  15. Sep 7, 2011 #14
    I am actually fascinated with things I don't understand. It makes me want to understand it! However not all are like me. For me, the easiest way for me to remember something is to associate it with weird stuff. I still remember stuff from my logic class and my prof used wacky stuff that stuck with me. I know its a bit late...but this thread reminded me of what some person put during PF Chat and I like it. Next time you do this, try getting your students to do it.

  16. Sep 7, 2011 #15
  17. Sep 7, 2011 #16
    @Ivan: Both cases are really the same. That's why we are able to solve for z using the two simultaneous equations.

    I share your same passion for knowledge! I really love puzzles...the thrill from solving them equals nothing else...And hey, thanks for sharing the Beautiful Dance Moves. That's really a good way to memorise the graphs! :D
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