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Difference between revisions of "1966 AHSME Problems/Problem 38"

(Created page with "== Problem == In triangle <math>ABC</math> the medians <math>AM</math> and <math>CN</math> to sides <math>BC</math> and <math>AB</math>, respectively, intersect in point <math>O<...")
 
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== Solution ==
 
== Solution ==
  
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Construct triangle <math>\triangle ABC</math> with points <math>M,N,P</math> being the midpoints of sides <math>\overline{CB}, \overline{AB}, \overline{AC}</math>, respectively. Proceed by drawing all medians. Then draw all medians (so draw <math>\overline{AM}, \overline{BP}, \overline{CN}</math>). Next, draw line <math>\overline{PM}</math> and label <math>\overline{PM}</math>'s intersection with <math>\overline{CN}</math> as the point <math>Q</math>. From the problem, the area of <math>\triangle QMO</math> is <math>n</math>, but by vertical angles we know that <math>\angle QOM = \angle AOM</math>. Furthermore, since line <math>\overline{PM}</math> is drawn from the midpoint of <math>\overline{AC}</math> to the midpoint of <math>\overline{CB}</math>, we know that <math>\overline{PM}</math> is parallel to <math>\overline{AB}</math> (via SAS similarity on triangles PCM and ABC). From these parallel lines we know that <math>\angle PMO = \angle OAB</math> which indicates that <math>\triangle QOM \sim \triangle ANO</math>. The linear ratio from <math>\triangle QOM</math> to <math>\triangle ANO</math> is 1:2 because line segment <math>\overline{OM}</math> is one half of line segment <math>\overline{AO}</math> since <math>\overline{AO}</math> and <math>\overline{OM}</math> make up the median <math>\overline{AM}</math>. Thus the area ratio is 1:4. So <math>\triangle ANO</math> has area <math>4n</math>. Since <math>\triangle ONB</math> has the same height and base as <math>\triangle ANO</math> we know that the area of <math>\triangle AOB = 8n</math>. The medians form 3 triangles each with area <math>1/3</math> of the total triangle (these triangles are <math>\triangle AOB, \triangle COB, \triangle COA</math>). Thus since <math>\triangle AOB = 4n \underset{\text{multiply by 3}}{\implies} \triangle ABC = 24n</math> <math>\fbox{D}</math>.
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- LJ
  
 
== See also ==
 
== See also ==

Latest revision as of 19:21, 3 April 2023

Problem

In triangle $ABC$ the medians $AM$ and $CN$ to sides $BC$ and $AB$, respectively, intersect in point $O$. $P$ is the midpoint of side $AC$, and $MP$ intersects $CN$ in $Q$. If the area of triangle $OMQ$ is $n$, then the area of triangle $ABC$ is:

$\text{(A) } 16n \quad \text{(B) } 18n \quad \text{(C) } 21n \quad \text{(D) } 24n \quad \text{(E) } 27n$

Solution

Construct triangle $\triangle ABC$ with points $M,N,P$ being the midpoints of sides $\overline{CB}, \overline{AB}, \overline{AC}$, respectively. Proceed by drawing all medians. Then draw all medians (so draw $\overline{AM}, \overline{BP}, \overline{CN}$). Next, draw line $\overline{PM}$ and label $\overline{PM}$'s intersection with $\overline{CN}$ as the point $Q$. From the problem, the area of $\triangle QMO$ is $n$, but by vertical angles we know that $\angle QOM = \angle AOM$. Furthermore, since line $\overline{PM}$ is drawn from the midpoint of $\overline{AC}$ to the midpoint of $\overline{CB}$, we know that $\overline{PM}$ is parallel to $\overline{AB}$ (via SAS similarity on triangles PCM and ABC). From these parallel lines we know that $\angle PMO = \angle OAB$ which indicates that $\triangle QOM \sim \triangle ANO$. The linear ratio from $\triangle QOM$ to $\triangle ANO$ is 1:2 because line segment $\overline{OM}$ is one half of line segment $\overline{AO}$ since $\overline{AO}$ and $\overline{OM}$ make up the median $\overline{AM}$. Thus the area ratio is 1:4. So $\triangle ANO$ has area $4n$. Since $\triangle ONB$ has the same height and base as $\triangle ANO$ we know that the area of $\triangle AOB = 8n$. The medians form 3 triangles each with area $1/3$ of the total triangle (these triangles are $\triangle AOB, \triangle COB, \triangle COA$). Thus since $\triangle AOB = 4n \underset{\text{multiply by 3}}{\implies} \triangle ABC = 24n$ $\fbox{D}$.

- LJ

See also

1966 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 37 		Followed by
Problem 39
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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