Question #158467

In a quadrilateral OABC, D is the midpoint of BC and E is the point on AD such that AE : ED = 2 : 1. Given that OA = A, OB = B, and OC = c express OD and OE in terms of a,b and c.


1
Expert's answer
2021-02-01T12:00:23-0500

The question does not contain complete information that is needed to resolve it. Depending on angles BOC\angle BOC and AOB\angle AOB , the answer may be different. If we change these angles and and all conditions are done, OD and OE change. It is shown on the picture


Solution:

Introduce vectors: OA=a\overrightarrow{OA}=\overrightarrow{a} , OB=b\overrightarrow{OB}=\overrightarrow{b} , OC=c\overrightarrow{OC}=\overrightarrow{c} .

Then OD=b+c2\overrightarrow{OD}=\frac{\overrightarrow{b}+\overrightarrow{c}}{2} (used BD = DC).

DA=aOD=ab+c2=2a(b+c)2\overrightarrow{DA}=\overrightarrow{a}-\overrightarrow{OD}=\overrightarrow{a}-\frac{\overrightarrow{b}+\overrightarrow{c}}{2}=\frac{2\overrightarrow{a}-(\overrightarrow{b}+\overrightarrow{c})}{2} .

AD and AM are medians. E is the point of the intersection of AD and AM. In this point each median divides as 2:1 (AE:AD = 2:1). That's why

DE=13DA=2a(b+c)6\overrightarrow{DE}=\frac13\overrightarrow{DA}=\frac{2\overrightarrow{a}-(\overrightarrow{b}+\overrightarrow{c})}{6}

OE=OD+DE=b+c22a(b+c)2=a+b+c3\overrightarrow{OE}=\overrightarrow{OD}+\overrightarrow{DE} =\frac{\overrightarrow{b}+\overrightarrow{c}}{2}-\frac{2\overrightarrow{a}-(\overrightarrow{b}+\overrightarrow{c})}{2}=\frac{\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}}{3}

OD=OD=b+c2=12(b+c)2=12b2+c2+2bccosBOCOD=|\overrightarrow{OD}|=|\frac{\overrightarrow{b}+\overrightarrow{c}}{2}|=\frac12\sqrt{(\overrightarrow{b}+\overrightarrow{c})^2}=\frac12\sqrt{b^2+c^2+2bc\cos{\angle{BOC}}}

OE=OE=a+b+c3=13(a+b+c)2=OE=|\overrightarrow{OE}|=|\frac{\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}}{3}|=\frac13\sqrt{(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c})^2}=

=13a2+b2+c2+2(abcosAOB+bccosBOC+accosAOC)=\frac13\sqrt{a^2+b^2+c^2+2(ab\cos{\angle{AOB}}+bc\cos{\angle{BOC}}+ac\cos{\angle{AOC}})}

An answer depends on angles BOC\angle BOC , AOB\angle AOB and AOC=BOC+AOB\angle AOC = \angle BOC+ \angle AOB.

We are free to change them without breaking any condition (as shown on the picture). So these angles cannot be expressed in terms of a, b and c.


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