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The point E on AB=AC - JBMO Shortlist

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#1 h Oct 30, 2010, 8:24 PM • 2 Y

Y by Adventure10, Mango247

Consider a triangle $ABC$ with $AB=AC$ , and $D$ the foot of the altitude from the vertex $A$ . The point $E$ lies on the side $AB$ such that $\angle ACE= \angle ECB=18^{\circ}$ .

If $AD=3$ , find the length of the segment $CE$ .

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#2 h Oct 31, 2010, 12:31 AM • 1 Y

Y by Adventure10

Lemma. $\cos{36} = \frac{1 + \sqrt{5}}{4}$ .

To prove this we draw a $36$ - $72$ - $72$ isosceles triangle. Let $A$ be the vertex angle and $B$ and $C$ be the base angles. Draw angle bisector $BD$ . Call $BC = a$ , $CD = b$ , and $AD = c$ ; then we see that $AB = b + c$ . So since $\triangle BDC \sim \triangle ACB$ , we have that $\frac{a}{b} = \frac{b + c}{a}$ . Also, by the Angle Bisector Theorem, $\frac{a}{b + c} = \frac{b}{c}$ , so $\frac{b + c}{a} = \frac{c}{b}$ . This means that $\frac{a}{b} = \frac{c}{b}$ if we substitute using our first proportion, so $a = c$ .

By the Law of Cosines on $\triangle BDC$ , we see that $b^2 = 2a^2 - 2a^2\cos{36}$ . Also, from the first propotion, $\frac{a}{b} = \frac{a + b}{a}$ . So $a^2 = ab + b^2 \Longrightarrow b^2 + ab - a^2 = 0$ . Hence, $b = \frac{-a \pm \sqrt{5a^2}}{2} = a \cdot \frac{-1 + \sqrt{5}}{2}$ .

Now, we plug this into the equation $b^2 = 2a^2 - 2a^2\cos{36}$ . $a^2 \cdot \frac{3 - \sqrt{5}}{2} = 2a^2 - 2a^2\cos{36} \Longrightarrow 3 - \sqrt{5} = 4 - 4\cos{36}$ . So $\cos{36} = \frac{1 + \sqrt{5}}{4}$ .

So from here we use Law of Cosines and solve. (I'm figuring out exactly how to proceed from here.)

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#3 h Nov 10, 2010, 1:33 AM • 8 Y

Y by Mathlover_1, Nomia, Adventure10, Mango247, Umudlu, FeridMath, Aykhan_Abbasli, AylyGayypow009

Let $AD$ intersect $CE$ at $I$ , and extend $AD$ to $F$ such that $BACF$ is a rhombus. Then angle chasing, we have $\angle AEI=\angle EAI=\angle IFC=\angle ICF=54$ , so $IA=IE$ and $IC=IF\implies CE=AF=2AD=6$ .

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