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Difference between revisions of "1960 IMO Problems/Problem 4"

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Construct the line <math>l_3</math> parallel to <math>l_2</math> so that the distance between <math>l_2</math> and <math>l_3</math> is <math>h_b</math> and <math>M_a</math> lies between these lines. <math>B</math> lies on <math>l_3</math>. Then <math>B=l_1\cap l_3</math>.
 
Construct the line <math>l_3</math> parallel to <math>l_2</math> so that the distance between <math>l_2</math> and <math>l_3</math> is <math>h_b</math> and <math>M_a</math> lies between these lines. <math>B</math> lies on <math>l_3</math>. Then <math>B=l_1\cap l_3</math>.
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==Video Solution==
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https://youtu.be/M0_UdvxH890
  
 
==See Also==
 
==See Also==
  
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{{IMO box|year=1960|num-b=3|num-a=5}}
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{{IMO7 box|year=1960|num-b=3|num-a=5}}
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[[Category:Olympiad Geometry Problems]]
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[[Category:Geometric Construction Problems]]

Latest revision as of 00:43, 2 January 2023

Problem

Construct triangle $ABC$, given $h_a$, $h_b$ (the altitudes from $A$ and $B$), and $m_a$, the median from vertex $A$.

Solution

Let $M_a$, $M_b$, and $M_c$ be the midpoints of sides $\overline{BC}$, $\overline{CA}$, and $\overline{AB}$, respectively. Let $H_a$, $H_b$, and $H_c$ be the feet of the altitudes from $A$, $B$, and $C$ to their opposite sides, respectively. Since $\triangle ABC\sim\triangle M_bM_aC$, with $M_bM_a=\frac12 AB$, the distance from $M_a$ to side $\overline{AC}$ is $\frac{h_b}{2}$.

Construct $AM_a$ with length $m_a$. Draw a circle centered at $A$ with radius $h_a$. Construct the tangent $l_1$ to this circle through $M_a$. $\overline{BC}$ lies on $l_1$.

Draw a circle centered at $M_a$ with radius $\frac{h_b}{2}$. Construct the tangent $l_2$ to this circle through $A$. $\overline{AC}$ lies on $l_2$. Then $C=l_1\cap l_2$.

Construct the line $l_3$ parallel to $l_2$ so that the distance between $l_2$ and $l_3$ is $h_b$ and $M_a$ lies between these lines. $B$ lies on $l_3$. Then $B=l_1\cap l_3$.

Video Solution

https://youtu.be/M0_UdvxH890

See Also

1960 IMO (Problems)
Preceded by
Problem 3 		1 • 2 • 3 • 4 • 5 • 6 • 7 Followed by
Problem 5
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