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.}}
