An extremely tough maths question

Test Preparation SAT Preparation
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<p>I have been trying to solve a question i read in this forum a long time ago but haven't been able to find the answer. Any ideas will be appreciated. :)</p>

<p>Given a triangle ABC as described below: Side AB = Side AC. Draw a line from C to side AB. call that line CD. Now draw a line from B to side AC. Call that line BE. Let angle EBC = 60 degrees, angle BCD equal 70 degrees, angle ABE equal 20 degrees, and angle DCE equal 10 degrees. Now draw line DE. The question--find what angle EDC is.</p>

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<p>Is the answer 70?</p>

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<p>Hmm, are you sure you’ve given all of the necessary info?</p>

<p>I hit a road block after a bit.</p>

4

<p>There is a contradiction in this problem. ABC has to be equal ABE + EBC = 20+60=80. Then, BAC=BCA (isosceles triangle) = (180-ABC)/2 = 100/2=50. </p>

<p>Since DCE=10 and BCD=70, BCA should be 10+70=80</p>

<p>Contradiction or misstated problem.</p>

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<p>YAHA, I don’t get what you’re saying - it is an isosceles triangle because ACB = ABC.</p>

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<p>Yes, thank you for pointing out. :wink:
First of all, there is no chance in the world that something like this will be on SAT. Second of all, it appears that some piece of information is missing from the problem :/</p>

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<p>HAHAHAHAHA!</p>

<p>This is actually a classic question, but it has nothing to do with the SAT.</p>

<p>[Gary</a> Gruber, Ph.D. - nationally recognized expert in the field of test preparation](<a href=“drgarygruber.com”>drgarygruber.com)</p>

<p>There are a few ways to do it… but all of them require all kinds of constructions and good knowledge of triangle similarity/congruence.</p>

<p>I’d suggest you google it. There are a few interesting solutions to be found.</p>

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<p>[A</a> hard Geometry problem for the CC Math geniuses!](<a href=“http://www.collegeconfidential.com/discus/messages/69/80505.html?]A”>http://www.collegeconfidential.com/discus/messages/69/80505.html?)</p>

<p>First posted on Tuesday, July 20, 2004. The solution is at the end of the thread.</p>

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<p>[url=<a href=“http://blogs.sun.com/simford/entry/solved_the_world_s_hardest]相忘于江湖[/url”>http://blogs.sun.com/simford/entry/solved_the_world_s_hardest]相忘于江湖[/url</a>] is another way. The problem in that page is slightly different letter-wise, but solving it uses the same process.</p>