Inequalities In Triangles / Discussion

This is the discussion related to the wiki page Inequalities In Triangles.

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Misstated conclusion in problem 94 page 41

Anonymous (97.87.15.x) 05 Oct 2008 00:10

I find the sum of segments connecting a point inside a triangle with its vertices is GREATER than the semiperimeter of the triangle.

A proof and several geometer's sketchpad experiments.

Re: Misstated conclusion in problem 94 page 41

Sumizdat 06 Oct 2008 22:08

You are right, of course — thank you! The correct formulation in the 1st edition of the book (1892) reads: Prove that: The sum of segments connecting a point inside a triangle with its vertices is smaller than the perimeter of the triangle, but greater than its semiperimeter.

In fact, the 2nd statement does not require the point to lie inside the triangle, while the 1st one merits the asterisk, warning of its difficulty level, and the Hint: See Figure 237.

Last edited on 07 Oct 2008 01:44 by Sumizdat Show more

Typo on p. 40, line 11 from bottom

Sumizdat 23 Oct 2009 23:19

$\angle BAD=\angle BAD''$ should be $\angle BAD=\angle DAB''$

Brent (guest) 17 Sep 2012 13:16

Misstated conclusion to problem 89 page 41
I find that a side of a triangle is LESS than its semiperimeter.

Sumizdat 22 Sep 2012 16:01

You are right (too :-). In fact, in an attempt to fix the error in problem 94 in the second printing of the book, the change was made in a wrong formulation, so that now there are errors in two problems, 89 and 94.

Exercise 86

blakester 01 Sep 2015 02:51

Exercise 86: Can an exterior angle of an isoceles triangle be smaller than the supplementary interior angle? Consider the cases when the angle is: (a) at the base, and (b) at the vertex.

(a)
$\S 42$ - An exterior angle of a triangle is greater than each interior angle not supplementary to it. $\S 35(2)$ - In an isoceles triangle, the angles at the base are congruent. The exterior angle at the base angle is greater than the angle congruent to its supplementary angle, so it cannot be smaller than the supplementary interior angle.

(b)
Let there be obtuse $\triangle ABC$ where the vertex $\angle B$ is obtuse. The exterior angle at $\angle B$ must be < $d$ ($\S 21 , \S 22$) and is less than the supplementary internal angle.

Exercise 87

blakester 01 Sep 2015 02:55

Exercise 87: Can a triangle have sides: (a) 1,2, and 3cm (centimeters) long? (b) 2,3, and 4cm long?

(a)
$\S 48$ - In a triangle, each side is smaller than the sum of the other two sides. $3 = 1 + 2$ would contradict this theorem.

(b)
$\S 48$, $2 < 3 + 4$ and $3 < 2 + 4$ and $4 < 2 + 3$ . Yes.

Exercise 88

blakester 01 Sep 2015 03:08

Exercise 88: Can a quadrilateral have sides: 2, 3, 4, and 10cm long?

No.
$\S 49$ - The line segment connecting any two points is less than any broken line connecting these points. $\S 32$ - A quadrilateral is a closed broken line and side of $10$ must be less than $2 + 3 + 4$ which it is not.

Exercise 89

blakester 01 Sep 2015 03:12

Exercise 89: A side of a triangle is smaller than its semiperimeter.

$\S 48$ - In a triangle, each side is smaller than the sum of the other two sides.
Given $\triangle ABC , AB < BC + AC$. Add $AB$ to both sides of the inequality. $AB + AB < AB + BC + AC$ or $2(AB) < AB + BC + AC$. Divide both sides of this inequality by 2 to produce the semi-perimeter on the right-hand side. $AB < \frac{AB + BC + AC}{2}$.

Exercise 90

blakester 01 Sep 2015 03:22

Exercise 90: A median of a triangle is smaller than its semiperimeter.

$BD < BC + CD$
$BD < AB + AD$
$BD + BD < AB + AD + BC + CD$
$2(BD) < AB + AD + BC + CD$

$AC = AD + DC$
$2(BD) < AB + BC + AC$
$BD < \frac{AB + BC + AC}{2}$

Exercise 91

blakester 01 Sep 2015 03:37

Exercise 91: A median drawn to a side of a triangle is smaller than the semisum of the other two sides.

Let there be a $\triangle ABC$ with median to side $AC$, $BD$. Continue median $BD$ past point $D$ a segment congruent to $BD$; label the terminal point $B'$. Connect points $A$ and $B'$ by segment to form $\triangle ADB'$.
$AD = \frac{AC}{2} = DC$; $\angle BDC = \angle ADB'$ ($\S 26$ - vertical angles); $DB' = BD$.
By $\S 40(1)$ SAS test, $\triangle ADB' = \triangle BDC$.
$BB' = 2(BD)$ <—— median to side $AC$. $\S 48$ - triangle inequality, $BB' < AB + AB'$
$2(BD) < AB + BC$
$BD < \frac{AB + BC}{2}$

Exercise 92

blakester 01 Sep 2015 03:55

Exercise 92: The sum of the medians of a triangle is smaller than its perimeter but greater than its semi-perimeter.

Let there be $\triangle ABC$. Median from $\angle A$ intersects $BC$ at midpoint $D$, median from $\angle B$ intersects $AC$ at midpoint $E$, median from $\angle C$ intersects $AB$ at midpoint $F$.

$\S 48$) In a triangle, each side is smaller than the sum of the other two sides.
Median $AD$ creates $\triangle ACD$ and by triangle inequality $AC < AD + CD$.
Median $BE$ creates $\triangle ABE$ and by triangle inequality $AB < BE + AE$.
Median $CF$ create $\triangle BCF$ and by triangle inequality $BC < CF + BF$.

$AB + AC + BC < BE + AE + AD + CD + CF + BF$
$AE = \frac{AC}{2}$ ; $CD = \frac{BC}{2}$ ; $BF = \frac{AB}{2}$
$AB + AC + BC < BE + \frac{AC}{2} + AD + \frac{BC}{2} + CF + \frac{AB}{2}$
$BE + AD + CF > \frac{AB + AC + BC}{2}$

The sum of the triangle medians is greater than the semi-perimeter.

Last edited on 13 Sep 2015 15:14 by blakester Show more

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