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

DAβ†’=aβ†’βˆ’ODβ†’=aβ†’βˆ’bβ†’+cβ†’2=2aβ†’βˆ’(bβ†’+cβ†’)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

DEβ†’=13DAβ†’=2aβ†’βˆ’(bβ†’+cβ†’)6\overrightarrow{DE}=\frac13\overrightarrow{DA}=\frac{2\overrightarrow{a}-(\overrightarrow{b}+\overrightarrow{c})}{6}

OEβ†’=ODβ†’+DEβ†’=bβ†’+cβ†’2βˆ’2aβ†’βˆ’(bβ†’+cβ†’)2=aβ†’+bβ†’+cβ†’3\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=∣ODβ†’βˆ£=∣bβ†’+cβ†’2∣=12(bβ†’+cβ†’)2=12b2+c2+2bccos⁑∠BOCOD=|\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=∣OEβ†’βˆ£=∣aβ†’+bβ†’+cβ†’3∣=13(aβ†’+bβ†’+cβ†’)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(abcos⁑∠AOB+bccos⁑∠BOC+accos⁑∠AOC)=\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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