1988 OIM Problems/Problem 1

Problem

The measurements of the sides of a triangle are in arithmetic progression and the lengths of the heights of the same triangle are also in arithmetic progression. Prove that the triangle is equilateral.

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution 1

{{Step 1: Let the sides of the triangle be a a, b b, and c c. Since the sides are in arithmetic progression, we can represent them as:

b = a + c 2 . b= 2 a+c ​

.

So, the three sides are a a, b = a + c 2 b= 2 a+c ​

, and 

c c, where a a and c c are the first and third terms of the arithmetic progression.

Step 2: Let the heights corresponding to these sides be h a h a ​

, 

h b h b ​

, and 

h c h c ​

.

We are also given that the heights are in arithmetic progression. So, we can represent the heights as:

h b = h a + h c 2 . h b ​

= 

2 h a ​

+h 

c ​

​

.

Step 3: Use the formula for the area of the triangle. The area A A of a triangle can be expressed using any side and the corresponding height:

A = 1 2 × side × corresponding height . A= 2 1 ​

×side×corresponding height.

Thus, we have the following three expressions for the area of the triangle:

A = 1 2 a h a = 1 2 b h b = 1 2 c h c . A= 2 1 ​

ah 

a ​

= 

2 1 ​

bh 

b ​

= 

2 1 ​

ch 

c ​

.

From these, we can express the heights in terms of the area and the side lengths:

h a = 2 A a , h b = 2 A b , h c = 2 A c . h a ​

= 

a 2A ​

,h 

b ​

= 

b 2A ​

,h 

c ​

= 

c 2A ​

.

Step 4: Use the fact that the heights are in arithmetic progression. Since the heights are in arithmetic progression, we know that:

h b = h a + h c 2 . h b ​

= 

2 h a ​

+h 

c ​

​

.

Substitute the expressions for h a h a ​

, 

h b h b ​

, and 

h c h c ​

 into this equation:

2 A b = 2 A a + 2 A c 2 . b 2A ​

= 

2 a 2A ​

+ 

c 2A ​

​

.

Simplify both sides:

2 A b = 2 A 2 ( 1 a + 1 c ) = A ( 1 a + 1 c ) . b 2A ​

= 

2 2A ​

( 

a 1 ​

+ 

c 1 ​

)=A( 

a 1 ​

+ 

c 1 ​

).

Canceling A A from both sides (assuming A ≠ 0 A  =0):

2 b = 1 a + 1 c . b 2 ​

= 

a 1 ​

+ 

c 1 ​

.

Step 5: Substitute the relation b = a + c 2 b= 2 a+c ​

.

We know that b = a + c 2 b= 2 a+c ​

. Substituting this into the equation above:

2 a + c 2 = 1 a + 1 c . 2 a+c ​

2 ​

= 

a 1 ​

+ 

c 1 ​

.

Simplifying the left-hand side:

4 a + c = 1 a + 1 c . a+c 4 ​

= 

a 1 ​

+ 

c 1 ​

.

Now, combine the terms on the right-hand side:

1 a + 1 c = a + c a c . a 1 ​

+ 

c 1 ​

= 

ac a+c ​

.

Thus, the equation becomes:

4 a + c = a + c a c . a+c 4 ​

= 

ac a+c ​

.

Step 6: Cross-multiply to simplify. Cross-multiply to eliminate the fractions:

4 a c = ( a + c ) 2 . 4ac=(a+c) 2

.

Expanding both sides:

4 a c = a 2 + 2 a c + c 2 . 4ac=a 2

+2ac+c 

2

.

Rearrange the terms:

4 a c − 2 a c = a 2 + c 2 , 4ac−2ac=a 2

+c 

2

,

2 a c = a 2 + c 2 . 2ac=a 2

+c 

2

.

Step 7: Recognize the equation. We now have the equation:

2 a c = a 2 + c 2 . 2ac=a 2

+c 

2

.

This is a well-known equation that holds if and only if a = c a=c, meaning the triangle is isosceles with a = c a=c.

Step 8: Conclude that the triangle is equilateral. If a = c a=c, then from the relation b = a + c 2 b= 2 a+c ​

, we see that 

b = a = c b=a=c. Therefore, all three sides of the triangle are equal, which means the triangle is equilateral.

Conclusion: We have shown that if the sides and the heights of a triangle are both in arithmetic progression, the triangle must be equilateral.}}

See also

https://www.oma.org.ar/enunciados/ibe3.htm