Maths Olympiad Prep

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Combinatorics Difficulty 5.1 AIME, harder Find the answer

4. If one focus of the ellipse is the orthocenter of the triangle formed by its three vertices, then the eccentricity of the ellipse e=e= \qquad

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Solution

4. 512\frac{\sqrt{5}-1}{2}.

Let F(c,0)F(c, 0) be a focus of the ellipse, which is the orthocenter of the triangle formed by the three vertices A(a,0)B(0,b)C(0,b)A(a, 0)、 B(0, b)、 C(0,-b) of the ellipse, as shown in Figure 5. From CFABC F \perp A B we have
bc(ba)=1b2=aca2c2a2=cae2+e1=0e=512. \begin{array}{l} \frac{b}{c}\left(-\frac{b}{a}\right)=-1 \\ \Rightarrow b^{2}=a c \\ \Rightarrow \frac{a^{2}-c^{2}}{a^{2}}=\frac{c}{a} \\ \Rightarrow e^{2}+e-1=0 \\ \Rightarrow e=\frac{\sqrt{5}-1}{2} . \end{array}

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.