bonanova Posted September 15, 2008 Report Share Posted September 15, 2008 (edited) This one is a lot like the first, but it's simpler. Triangle ABC is isosceles. Find angle x. [edit - give the solution] Do not assume the figure is drawn to scale. Do not use the law of sines. Use simple geometric reasoning. Since it's easier, you may draw only one auxiliary line. Have fun. Edited September 15, 2008 by bonanova Added "give the solution" Quote Link to comment Share on other sites More sharing options...
0 Guest Posted September 15, 2008 Report Share Posted September 15, 2008 (edited) 20 again? my points EB_=30 DA_=20 A_D=110 AD_=50 bahh running out of time type the rest out later Edited September 15, 2008 by solemnraven Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted September 15, 2008 Author Report Share Posted September 15, 2008 20 again? Nope. Different angles. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted September 15, 2008 Report Share Posted September 15, 2008 A_B=70 B_E=110 E_D=70 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted September 15, 2008 Report Share Posted September 15, 2008 so now we have to figure out what x is and how this could possibly be easier than the last one... 30 degrees? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted September 15, 2008 Report Share Posted September 15, 2008 so now we have to figure out what x is and how this could possibly be easier than the last one... 30 degrees? That was my first thought. traingle ADE is similar to trangle CDB, but I can't prove it...yet Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted September 16, 2008 Author Report Share Posted September 16, 2008 so now we have to figure out what x is and how this could possibly be easier than the last one... 30 degrees? It might be. Can you show why? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted September 16, 2008 Report Share Posted September 16, 2008 I think I have it but it will be hard to explain. 40 degrees. I worked out as many angles as I could using the fact that the internal angles of a triangle add to 180. And that opposite angles of a pair of crossing lines are equal. That left me with x and the three other angles adjacent to the line DE. The angles at points E and D must each equal 180 as must the internal angles of the triangle which icludes x. I could then see that x=CDE-20, EDB=CED-30, and CDE=x+20. If the x triangle is an isosceles, then EDB is seventy and x is 40. Plugging those values in gave consistant answers. Varying x to 30 gives two results for CDE and EDB. Only 40 degrees holds true. Sorry for the clumsy explanation. I left out one step but it was getting so long winded. Quote Link to comment Share on other sites More sharing options...
0 soop Posted September 16, 2008 Report Share Posted September 16, 2008 I'm a bit stuck, can someone help please? PM me or post a spoiler if you do! I've drawn out the angles for the 3 bottom most triangles, and the bottom angle of the triangle X is in. What is meant by auxilary line? I'm guessing it can't be done without this, but do I get to draw a line of a known angle through the triangle? Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted September 16, 2008 Author Report Share Posted September 16, 2008 I'm a bit stuck, can someone help please? PM me or post a spoiler if you do!I've drawn out the angles for the 3 bottom most triangles, and the bottom angle of the triangle X is in. What is meant by auxilary line? I'm guessing it can't be done without this, but do I get to draw a line of a known angle through the triangle? The solution to the first puzzle used two auxiliary lines. For example. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted September 16, 2008 Report Share Posted September 16, 2008 Got it!!!! I feel relieved now... The answer is 30. I'll try to prove it without a drawing, but I'm sure bonanova will eventually post the drawing. If we draw a line from B but at 60 degrees instead of 50, it crosses first with AE at, let's call it M and with AC at let's call it F. Angle ADB=50 (trivial), thus triangle ADB is isosceles, so AD=AB. Triangle AMB is equilateral, so AB=AM. So AD=AM, thus ADM is isosceles. Since angle DAM is 20, angles ADM and AMD are 80. Thus angle DMF is 180-BMA-AMD=40. But DFM is also 40 (trivial) so DFM is isosceles. But also FME is equilateral (trivial). Therefore the quadrilateral DFEM is symmetric, and the axis of symmetry is DE. Therefore x is the bisector of MEF (60deg), thus it's equal to 30. Great puzzle, 5 stars! Quote Link to comment Share on other sites More sharing options...
0 Guest Posted September 17, 2008 Report Share Posted September 17, 2008 Got it!!!! I feel relieved now... The answer is 30. I'll try to prove it without a drawing, but I'm sure bonanova will eventually post the drawing. If we draw a line from B but at 60 degrees instead of 50, it crosses first with AE at, let's call it M and with AC at let's call it F. Angle ADB=50 (trivial), thus triangle ADB is isosceles, so AD=AB. Triangle AMB is equilateral, so AB=AM. So AD=AM, thus ADM is isosceles. Since angle DAM is 20, angles ADM and AMD are 80. Thus angle DMF is 180-BMA-AMD=40. But DFM is also 40 (trivial) so DFM is isosceles. But also FME is equilateral (trivial). Therefore the quadrilateral DFEM is symmetric, and the axis of symmetry is DE. Therefore x is the bisector of MEF (60deg), thus it's equal to 30. Great puzzle, 5 stars! Nice Proof Kudos to you Quote Link to comment Share on other sites More sharing options...
0 Guest Posted September 17, 2008 Report Share Posted September 17, 2008 Nice Proof Kudos to you And the drawing Quote Link to comment Share on other sites More sharing options...
0 Guest Posted September 17, 2008 Report Share Posted September 17, 2008 I thought you could only add 1 new line? When I first said 30 degrees, I didn't have any proof, but i had some hunch that the unknown angle CDE was equal to angle DA_. I just couldn't figure out why. And if CDE was 50, then ED_ had to be 80, which leaves x at 30. Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted September 17, 2008 Author Report Share Posted September 17, 2008 Very nice. And thanks to Ben Law for the figure. Marinja correctly points out an error in the OP which restricts the solution to a single extra line. There is a simpler construction, but it instead adds a single extra point. And that point defines two extra lines. So to my knowledge a single extra line won't do it. My bad. somewhere on line BC.despistado's explanation is formalized, the last two lines could be rewritten ]Therefore the quadrilateral DFEM is symmetric, and the axis of symmetry is DE. Therefore x is the bisector of MEF (60deg), thus it's equal to 30. Since DF=DM, FE=ME and of course DE=DE, triangles DEF and DEM have three equal sides [sSS] and are congruent [mirror images]. Therefore angle FED = angle MED. Since their sum is 60, they both equal 30. Quote Link to comment Share on other sites More sharing options...
0 soop Posted September 17, 2008 Report Share Posted September 17, 2008 So to my knowledge a single extra line won't do it. My bad. O___O; Ah, never mind, I probably wouldn't have got it anyway And you're not the only one to make a mistake.. >_> Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted September 17, 2008 Author Report Share Posted September 17, 2008 So to my knowledge a single extra line won't do it. My bad. O___O; Ah, never mind, I probably wouldn't have got it anyway And you're not the only one to make a mistake.. >_> The solution with one extra point hasn't been found yet. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted September 18, 2008 Report Share Posted September 18, 2008 The solution with one extra point hasn't been found yet. Eureka! If we trace a segment that forms 20 degrees with AB until it crosses BC, that point is called P. And now the deductions: Angle ADB=50º (trivial) thus ADB is isosceles and AD=AB. Angle APB=80º (trivial) thus APB is isosceles and AP=AB. Thus AD=AP. But Angle DAP=60º, so DAP is equilateral, thus DP=AD=AP. Both angles AEP and EAP are 40º (trivial) so AEP is isosceles thus AP=PE. Thus DP=PE, therefore PDE is isosceles. Since Angle DPE is 180-60-80=40, and PDE is isosceles, angle DEP has to be 70. Thus x=70-40=30. Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted September 18, 2008 Author Report Share Posted September 18, 2008 Yup. Quote Link to comment Share on other sites More sharing options...
Question
bonanova
This one is a lot like the first, but it's simpler.
Triangle ABC is isosceles.
Find angle x. [edit - give the solution]
Do not assume the figure is drawn to scale.
Do not use the law of sines.
Use simple geometric reasoning.
Since it's easier, you may draw only one auxiliary line.
Have fun.
Edited by bonanovaAdded "give the solution"
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