Find the length of AB [KMO #4, Grade 9]

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Find the length of AB [KMO #4, Grade 9]

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Pythagorean Theorem

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Obel1x

#1 h Mar 13, 2011, 8:18 PM • 2 Y

Y by Adventure10, Mango247

In triangle $ABC$ medians of triangle $BE$ and $AD$ are perpendicular to each other. Find the length of $\overline{AB}$ , if $\overline{BC}=6$ and $\overline{AC}=8$

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#2 h Mar 13, 2011, 8:48 PM • 2 Y

Y by Adventure10, Mango247

Let the intersection of $AD,BE$ be $O$ . Let $OE=e,OD=D$ . Then $OA=2d,OB=2e$ .
By the Pythagorean Theorem, $e^2+4d^2=AE^2=16,d^2+4e^2=DB^2=9$ , adding which yields $d^2+e^2=5$ .
Thus, $AB=\sqrt{4d^2+4e^2}=\sqrt{4(d^2+e^2)}=\sqrt{4(5)}=2\sqrt{5}$ .

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#3 h Dec 30, 2019, 3:21 PM • 3 Y

Y by Ha_ha_ha, Adventure10, Mango247

BigSams wrote:

Let the intersection of $AD,BE$ be $O$ . Let $OE=e,OD=D$ . Then $OA=2d,OB=2e$ .
By the Pythagorean Theorem, $e^2+4d^2=AE^2=16,d^2+4e^2=DB^2=9$ , adding which yields $d^2+e^2=5$ .
Thus, $AB=\sqrt{4d^2+4e^2}=\sqrt{4(d^2+e^2)}=\sqrt{4(5)}=2\sqrt{5}$ .

Could you tell me why, $d^2+e^2=5$

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primesarespecial

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primesarespecial

#4 h Dec 22, 2020, 12:24 PM • 4 Y

Y by Ha_ha_ha, Mango247, Mango247, Mango247

@above Because the point where two medians intersect has to be the centroid and the centroid divides a median in the ratio 2:1

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#5 h Dec 22, 2020, 2:56 PM • 2 Y

Y by Ha_ha_ha, Mango247

I am going to BARY this problem.

Use Barycentrics on $\triangle ABC$ and set $A=(1,0,0),B=(0,1,0),C=(0,0,1),D=(0,\tfrac12,\tfrac12),E=(\tfrac12,0,\tfrac12)$ . We have $\overrightarrow{AD}=(-1,\tfrac12,\tfrac12)$ and $\overrightarrow{BE}=(\tfrac12,-1,\tfrac12)$ . By EFFT, $\overrightarrow{AD}\perp\overrightarrow{BE} \iff 6^2(\tfrac14-\tfrac12)+8^2(-\tfrac12+\tfrac14)+c^2(1+\tfrac14)=0$ . Solving for $c$ , our answer is $2\sqrt5$ . $\blacksquare$

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#6 h Apr 30, 2023, 12:09 PM

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Let $O$ be the intersection of $AD,BE$
Let $OE=a$ and $OD=b$
$\Rightarrow OB=2a, OA=2b$
Using pythagorean theorem, we get:
$a^2+(2b)^2=AE^2 \Rightarrow a^2+4b^2=4^2=16$
$b^2+(2a)^2=BD^2 \Rightarrow b^2+4a^2=3^2=9$
Adding them together, we get:
$5a^2+5b^2=25 \Rightarrow a^2+b^2=5$
Finally, we get:
$AB^2=(2a)^2+(2b)^2=4a^2+4b^2=4(a^2+b^2)=4 \cdot 5=20 \Rightarrow AB= \sqrt{20} =2 \sqrt{5}$ $\blacksquare$

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