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# Geometry

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Some time ago I was working on this geometry exercise, I ended up with a solution that was quite long and delved into sine equations. I wonder if anyone can think up a shorter solution.

Problem:

There is a triangle ABC, the bisector, the altitude and the median divide angle ACB in four equal parts. Find all possible angels angle ACB can be. This needs to be solved without help from software like the calculator.

The answer is 90 degrees but I wonder if this answer can be achieved without using sine calculation. Perhaps proof by contradiction is an option?

Honga  Aug 31, 2014

#6
+18827
+10

I called each of the four small angles at C, also  $$\alpha$$

$$\boxed { \cos{(\alpha)} = \frac{h}{b} = \frac{h}{d} \quad | \quad h=\overline{EC} \quad \mbox{and} \quad d = \overline{FC} } \Rightarrow \boxed{d=b}$$

$$Areas: DCB = DCA = \frac{ \frac{c}{2} *h }{ 2 }$$

I.   Areas: ACF + FCD = DCB

$$b^2\sin{(2\alpha)} +be\sin{(\alpha)} =ae\sin{(\alpha)}$$

II.  Areas: FCD + DCB = FCB

$$be\sin{(\alpha)} +ae\sin{(\alpha)} =ab\sin{(2\alpha)}$$

I.+II.:

$$\begin{array}{rcl} b^2\sin{(2\alpha)} +2be\sin{(\alpha)} &=&ab\sin{(2\alpha)} \quad | \quad :b\\ b\sin{(2\alpha)} +2e\sin{(\alpha)} &=& a\sin{(2\alpha)} \end{array}$$

$$\Rightarrow \boxed{ \frac { 2e\sin{( \alpha) } } { \sin{(2\alpha) } } =a-b \qquad (1) }$$

I.-II.:

$$\begin{array}{rcl} b^2\sin{(2\alpha)} - ea\sin{(\alpha)} &=&ea\sin{(\alpha)} -ab\sin{(2\alpha)} \\ 2ea\sin{(\alpha)} &=& ab\sin{(2\alpha)} + b^2\sin{(2\alpha)} \end{array}$$

$$\Rightarrow \boxed{ \frac { 2e\sin{( \alpha) } } { \sin{(2\alpha) } } =\frac{ ab+b^2 } { a } \qquad (2) }$$

(1)=(2):

$$a-b =\frac{ ab+b^2 } { a }$$

$$\Rightarrow \boxed{ a^2 -2b*a -b^2 = 0 }$$

$$a_{1,2}={ 2b\pm\sqrt{4b^2-4*1*(-b^2)}\over2*1 }=b(1\pm\sqrt{2}) \quad | \quad \mbox{ b not negativ!}$$

$${ \boxed{ { a= b (1+\sqrt{2}) } } }$$

III. Areas: ACF + FCB = ACB

$$b^2\sin{(2\alpha)} + ba\sin{(2\alpha)} = ba\sin{(4\alpha)} \quad | \quad a=b(1+\sqrt{2})$$

$$b^2\sin{(2\alpha)} + b^2(1+\sqrt{2})\sin{(2\alpha)} = b^2(1+\sqrt{2})\sin{(4\alpha)} \quad | \quad :b^2$$

$$\sin{(2\alpha)} + (1+\sqrt{2})\sin{(2\alpha)} = (1+\sqrt{2})\sin{(4\alpha)}$$

$$(2+\sqrt{2})\sin{(2\alpha)} = (1+\sqrt{2})\sin{(4\alpha)} \quad | \quad \leftrightarrow$$

$$(1+\sqrt{2})\sin{(4\alpha)} = (2+\sqrt{2})\sin{(2\alpha)} \quad | \quad \sin{(4\alpha)}=2\sin{(2\alpha)}\cos{2\alpha}$$

$$(1+\sqrt{2})2\sin{(2\alpha)}\cos{2\alpha}= (2+\sqrt{2})\sin{(2\alpha)} \quad | \quad : \sin{(2\alpha)}$$

$$(1+\sqrt{2})2\cos{2\alpha}= (2+\sqrt{2})$$

$$(2+2\sqrt{2})\cos{2\alpha}= (2+\sqrt{2})$$

$$\cos{2\alpha}=\frac{ (2+\sqrt{2}) }{ (2+2\sqrt{2}) } \quad | \quad *\frac{ \frac{\sqrt{2}}{2} }{ \frac{\sqrt{2}}{2} }$$

$$\cos{2\alpha}=\frac{ (2+\sqrt{2})\frac{\sqrt{2}}{2} }{ (\sqrt{2}+2) } = \frac{ \sqrt{2} }{ 2 } \quad | \quad \pm \cos^{-1}()$$

$$2\alpha= \pm \cos^{-1}( \frac{ \sqrt{2} } { 2 } ) \quad | \quad {+} solution!$$

$$2\alpha= {+} \cos^{-1}( \frac{ \sqrt{2} } { 2 } ) =45\ensurement{^{\circ}} \quad | \quad *2$$

$${ \boxed{ { 4\alpha= 90\ensurement{^{\circ}} } } }$$

heureka  Sep 2, 2014
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#1
+5

The three lines include the altitude. From this, we know that ACE and EFC are right triangles. Since the next line is the median, and that the four parts are equal, we know that AE and EF are equal. EC is a shared segment. With SAS, these triangles are congruent, providing us with an isosoles triangle. The CD segment is the median, for my own reference while writing this answer. Anyway, we now know that segment a = CF. CF is the bisector.

Can you go on this?

Guest Aug 31, 2014
#2
+889
+5

The only method that I can see uses trig.

I called each of the four small angles at C, $$\alpha$$, and after five (easy) lines arrived at

$$\sin2\alpha\cos4\alpha=0,$$

implying that $$4\alpha=90$$, as required.

Bertie  Sep 1, 2014
#3
+169
0

Please show me the full proof, the main difficulty with this problem is the danger of falling for circular reasoning and I must be sure this is not the case in your reasoning. On another note the existence of the bisector in this triangle is completely irrelevant. The  only thing that constitutes this particular shape is the relationship of the median to the altitude.

Honga  Sep 1, 2014
#4
+889
+10

Label the four small angles at C as stated earlier, - use the sine rule, a/sinA=b/sinB etc, (remember that sin(90-P)=cosP) in the triangles ACD and ACB, - divide one equation by the other to remove the common sides (AC  and AD, AB(=2AD)), - cross multiply, - then use first the identity for sin(2P), and then the identity for sin(P+Q). That gets you to the equation I stated in my earlier post.

Post again if you have any problems.

Bertie  Sep 1, 2014
#5
+169
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To be honest the first part of your proof is exactly the same as the one I came up with. I think I messed up with the sine calculation though, my proof consisted of one and half page filled with sine equation reduction to get to a similar point you ended up with. Thanks for your help, +1!

Honga  Sep 2, 2014
#6
+18827
+10

I called each of the four small angles at C, also  $$\alpha$$

$$\boxed { \cos{(\alpha)} = \frac{h}{b} = \frac{h}{d} \quad | \quad h=\overline{EC} \quad \mbox{and} \quad d = \overline{FC} } \Rightarrow \boxed{d=b}$$

$$Areas: DCB = DCA = \frac{ \frac{c}{2} *h }{ 2 }$$

I.   Areas: ACF + FCD = DCB

$$b^2\sin{(2\alpha)} +be\sin{(\alpha)} =ae\sin{(\alpha)}$$

II.  Areas: FCD + DCB = FCB

$$be\sin{(\alpha)} +ae\sin{(\alpha)} =ab\sin{(2\alpha)}$$

I.+II.:

$$\begin{array}{rcl} b^2\sin{(2\alpha)} +2be\sin{(\alpha)} &=&ab\sin{(2\alpha)} \quad | \quad :b\\ b\sin{(2\alpha)} +2e\sin{(\alpha)} &=& a\sin{(2\alpha)} \end{array}$$

$$\Rightarrow \boxed{ \frac { 2e\sin{( \alpha) } } { \sin{(2\alpha) } } =a-b \qquad (1) }$$

I.-II.:

$$\begin{array}{rcl} b^2\sin{(2\alpha)} - ea\sin{(\alpha)} &=&ea\sin{(\alpha)} -ab\sin{(2\alpha)} \\ 2ea\sin{(\alpha)} &=& ab\sin{(2\alpha)} + b^2\sin{(2\alpha)} \end{array}$$

$$\Rightarrow \boxed{ \frac { 2e\sin{( \alpha) } } { \sin{(2\alpha) } } =\frac{ ab+b^2 } { a } \qquad (2) }$$

(1)=(2):

$$a-b =\frac{ ab+b^2 } { a }$$

$$\Rightarrow \boxed{ a^2 -2b*a -b^2 = 0 }$$

$$a_{1,2}={ 2b\pm\sqrt{4b^2-4*1*(-b^2)}\over2*1 }=b(1\pm\sqrt{2}) \quad | \quad \mbox{ b not negativ!}$$

$${ \boxed{ { a= b (1+\sqrt{2}) } } }$$

III. Areas: ACF + FCB = ACB

$$b^2\sin{(2\alpha)} + ba\sin{(2\alpha)} = ba\sin{(4\alpha)} \quad | \quad a=b(1+\sqrt{2})$$

$$b^2\sin{(2\alpha)} + b^2(1+\sqrt{2})\sin{(2\alpha)} = b^2(1+\sqrt{2})\sin{(4\alpha)} \quad | \quad :b^2$$

$$\sin{(2\alpha)} + (1+\sqrt{2})\sin{(2\alpha)} = (1+\sqrt{2})\sin{(4\alpha)}$$

$$(2+\sqrt{2})\sin{(2\alpha)} = (1+\sqrt{2})\sin{(4\alpha)} \quad | \quad \leftrightarrow$$

$$(1+\sqrt{2})\sin{(4\alpha)} = (2+\sqrt{2})\sin{(2\alpha)} \quad | \quad \sin{(4\alpha)}=2\sin{(2\alpha)}\cos{2\alpha}$$

$$(1+\sqrt{2})2\sin{(2\alpha)}\cos{2\alpha}= (2+\sqrt{2})\sin{(2\alpha)} \quad | \quad : \sin{(2\alpha)}$$

$$(1+\sqrt{2})2\cos{2\alpha}= (2+\sqrt{2})$$

$$(2+2\sqrt{2})\cos{2\alpha}= (2+\sqrt{2})$$

$$\cos{2\alpha}=\frac{ (2+\sqrt{2}) }{ (2+2\sqrt{2}) } \quad | \quad *\frac{ \frac{\sqrt{2}}{2} }{ \frac{\sqrt{2}}{2} }$$

$$\cos{2\alpha}=\frac{ (2+\sqrt{2})\frac{\sqrt{2}}{2} }{ (\sqrt{2}+2) } = \frac{ \sqrt{2} }{ 2 } \quad | \quad \pm \cos^{-1}()$$

$$2\alpha= \pm \cos^{-1}( \frac{ \sqrt{2} } { 2 } ) \quad | \quad {+} solution!$$

$$2\alpha= {+} \cos^{-1}( \frac{ \sqrt{2} } { 2 } ) =45\ensurement{^{\circ}} \quad | \quad *2$$

$${ \boxed{ { 4\alpha= 90\ensurement{^{\circ}} } } }$$

heureka  Sep 2, 2014

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