Difference between revisions of "1988 OIM Problems/Problem 1"

Latest revision as of 23:28, 19 March 2025

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

Let the sides be $a,a+d_1,a+2d_1$ , let the heights be $b,b-d_2,b-2d_2$ , and assume for the sake of contradiction that $d_1,d_2$ are both positive. Then, since longer heights must correspond to shorter bases (since multiplying yields equal areas) we must have $a$ correspond to $b$ , $a+d_1$ correspond to $b-d_2$ , and $a+2d_1$ correspond to $b-2d_2$ . Then, by triangle areas: $ab=(a+d_1)(b-d_2)=(a+2d_1)(b-2d_2)$ From the first equality, we find that $bd_1=ad_2+d_1d_2$ . From the equality $ab=(a+2d_1)(b-2d_2)$ , we derive $bd_1=ad_2+2d_1d_2$ . Substituting the first equation into the second results in $ad_2+d_1d_2=ad_2+2d_1d_2$ so $d_1d_2=0$ . But if only one of $d_1$ and $d_2$ is zero, then different heights (or bases) will correspond to the same base (or height), resulting in the same area, which is impossible. Thus $d_1=d_2=0$ , and the conclusion follows.

~ eevee9406

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Difference between revisions of "1988 OIM Problems/Problem 1"

Latest revision as of 23:28, 19 March 2025

Problem

Solution

See also

@@ Line 4: / Line 4: @@
 ~translated into English by Tomas Diaz. ~orders@tomasdiaz.com
-== Solution 1 ==
+==Solution==
-{{Step 1: Let the sides of the triangle be
+Let the sides be <math>a,a+d_1,a+2d_1</math>, let the heights be <math>b,b-d_2,b-2d_2</math>, and assume for the sake of contradiction that <math>d_1,d_2</math> are both positive. Then, since longer heights must correspond to shorter bases (since multiplying yields equal areas) we must have <math>a</math> correspond to <math>b</math>, <math>a+d_1</math> correspond to <math>b-d_2</math>, and <math>a+2d_1</math> correspond to <math>b-2d_2</math>. Then, by triangle areas:
-a
+<cmath>ab=(a+d_1)(b-d_2)=(a+2d_1)(b-2d_2)</cmath>
-a,
+From the first equality, we find that <math>bd_1=ad_2+d_1d_2</math>. From the equality <math>ab=(a+2d_1)(b-2d_2)</math>, we derive <math>bd_1=ad_2+2d_1d_2</math>. Substituting the first equation into the second results in
-b
+<cmath>ad_2+d_1d_2=ad_2+2d_1d_2</cmath>
-b, and
+so <math>d_1d_2=0</math>. But if only one of <math>d_1</math> and <math>d_2</math> is zero, then different heights (or bases) will correspond to the same base (or height), resulting in the same area, which is impossible. Thus <math>d_1=d_2=0</math>, and the conclusion follows.
-c
-c.
-Since the sides are in arithmetic progression, we can represent them as:
-b
+~ [https://artofproblemsolving.com/wiki/index.php/User:Eevee9406 eevee9406]
-=
-a
-+
-c
-.
-b=
-a+c
- .
-So, the three sides are
-a
-a,
-b
-=
-a
-+
-c
-b=
-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
-.
-h
-b
- =
-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
-=
-×
-side
-×
-corresponding height
-.
-A=
- ×side×corresponding height.
-Thus, we have the following three expressions for the area of the triangle:
-A
-=
-a
-h
-a
-=
-b
-h
-b
-=
-c
-h
-c
-.
-A=
- ah
-a
- =
- bh
-b
- =
- ch
-c
- .
-From these, we can express the heights in terms of the area and the side lengths:
-h
-a
-=
-A
-a
-,
-h
-b
-=
-A
-b
-,
-h
-c
-=
-A
-c
-.
-h
-a
- =
-a
-A
- ,h
-b
- =
-b
-A
- ,h
-c
- =
-c
-A
- .
-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
-.
-h
-b
- =
-h
-a
- +h
-c
- .
-Substitute the expressions for
-h
-a
-h
-a
- ,
-h
-b
-h
-b
- , and
-h
-c
-h
-c
-  into this equation:
-A
-b
-=
-A
-a
-+
-A
-c
-.
-b
-A
- =
-a
-A
- +
-c
-A
- .
-Simplify both sides:
-A
-b
-=
-A
-(
-a
-+
-c
-)
-=
-A
-(
-a
-+
-c
-)
-.
-b
-A
- =
-A
- (
-a
- +
-c
- )=A(
-a
- +
-c
- ).
-Canceling
-A
-A from both sides (assuming
-A
-≠
-A
-
-=0):
-b
-=
-a
-+
-c
-.
-b
- =
-a
- +
-c
- .
-Step 5: Substitute the relation
-b
-=
-a
-+
-c
-b=
-a+c
- .
-We know that
-b
-=
-a
-+
-c
-b=
-a+c
- . Substituting this into the equation above:
-a
-+
-c
-=
-a
-+
-c
-.
-a+c
- =
-a
- +
-c
- .
-Simplifying the left-hand side:
-a
-+
-c
-=
-a
-+
-c
-.
-a+c
- =
-a
- +
-c
- .
-Now, combine the terms on the right-hand side:
-a
-+
-c
-=
-a
-+
-c
-a
-c
-.
-a
- +
-c
- =
-ac
-a+c
- .
-Thus, the equation becomes:
-a
-+
-c
-=
-a
-+
-c
-a
-c
-.
-a+c
- =
-ac
-a+c
- .
-Step 6: Cross-multiply to simplify.
-Cross-multiply to eliminate the fractions:
-a
-c
-=
-(
-a
-+
-c
-)
-.
-ac=(a+c)
- .
-Expanding both sides:
-a
-c
-=
-a
-+
-a
-c
-+
-c
-.
-ac=a
- +2ac+c
- .
-Rearrange the terms:
-a
-c
-−
-a
-c
-=
-a
-+
-c
-,
-ac−2ac=a
- +c
- ,
-a
-c
-=
-a
-+
-c
-.
-ac=a
- +c
- .
-Step 7: Recognize the equation.
-We now have the equation:
-a
-c
-=
-a
-+
-c
-.
-ac=a
- +c
- .
-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
-b=
-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