This math problem owns you all

Colleges & Universities A-Z University of Pennsylvania
41

<p>This is a good problem. I'm close to the end. This is similar to a problem called the Langley Problem. The only difference is angle measures. Check out this website guys to see how it's done with slightly different numbers. <a href="http://agutie.homestead.com/files/LangleyProblem.html%5B/url%5D"&gt;http://agutie.homestead.com/files/LangleyProblem.html&lt;/a&gt;&lt;/p>

42

<p>The answer is 70.</p>

43

<p>The answer is 70.</p>

44

<p>This thread sucks</p>

45

<p>You're wrong pentasa, you suck.</p>

46

<p>Eh whatever, the answer's 20. There's only one right answer to this (try constructing it yourself, even). But the challenge is figuring out how it's 20.</p>

47

<p>it could be 110
every1 put it in the traingle see if it works
EDC=110
ADE=40
AED=120
DEB=20</p>

48

<p>it can't be.. between a 70 and a 60 degree angle, the 70 degree angle will project higher onto one side of the triangle, and the other one lower. therefore, the angle EDC must be less than the angle if it were parallel (70). so it really can't be 110. it seems to work mathematically, but it doesn't.. i'm still trying to figure out why it's 20, though..</p>

49

<p>Angles B and C are both 80. A would then be 20.</p>

50

<p>But if it's isosceles, then wouldn't angle C have to be equal to angle A?</p>

<p>mBA = mBC</p>

<p>so angle A = angle C ?</p>

51

<p>"Draw an isosceles triangle ABC with Side AB = Side AC"</p>

<p>AB = AC. Meaning angles C and B are the same in this triangle.</p>

52

<p>Heh, i did read it wrong :P</p>

53

<p>okay, so </p>

<p>ADE = w
ED(intersection) = x
DE(intersection) = y
AED = z</p>

<p>Using different angle sum properties you can get the following:</p>

<p>x + y = 130
z + y = 140
w + x = 150
w + z = 160</p>

<p>Unsolvable system of equations =/ Anyone see one im missing? I can't find a way to relate x and z/ :(</p>

54

<p>w+x+y+z = 290...
it doesn't help...but it exists...</p>

55

<p>tymps, yeah thats the exact same place I got stuck in</p>

56

<p>This can't be solved by systems like I tried at first. Look at this link that I posted before. It shows exactly how to do this problem with different angles. <a href="http://agutie.homestead.com/files/LangleyProblem.html%5B/url%5D"&gt;http://agutie.homestead.com/files/LangleyProblem.html&lt;/a>. I'd like to see the OP post a solution to the problem he claims "owns us all".</p>

57

<p>I didn't look into the problem but tymps, solving your system (by using rref of the matrix) leads to </p>

<p>w= 160-z
x= z-10
y= 140-z
z= z</p>

<p>Not sure if that helps</p>

58

<p>I can't say that that help incredibly much, dear...but thank you all the same</p>

59

<p>the method from the link doesn't really work. they create equilateral triangles etc. to solve that question. it doesn't work for this one unless i'm missing something</p>

60

<p>Why doesn't that work? Using equilateral triangles, etc. is using geometry like the OP said. The original set up it exactly the same I believe. An angle of 20 deg, and two of 80 deg. The only thing is in the problem on the link, the two angles are split up into 60&20 deg and 50&30 deg while the problem on CC has them split up as 60&20 deg and 70&10 deg. Besides that, it looks exactly the same to me.</p>