Inequalities In Triangles

page revision: 0, last edited: 15 Sep 2008 08:55

Inequalities In Triangles

page revision: 0, last edited: 15 Sep 2008 08:55

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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.

ReplyOptionsYou 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

insidethe triangle, while the 1st one merits the asterisk, warning of its difficulty level, and theHint:See Figure 237.ReplyOptions$\angle BAD=\angle BAD''$ should be $\angle BAD=\angle DAB''$

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

ReplyOptionsYou 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.

ReplyOptionsExercise 86:

(a)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.$\S 42$ - 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.

ReplyOptionsExercise 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.

ReplyOptionsExercise 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.

ReplyOptionsExercise 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. $AB < \frac{AB + BC + AC}{2}$.

ReplyOptionsExercise 90:

A median of a triangle is smaller than its semiperimeter.$BD < BC + CD$

$BD < AB + AD$

$BD + BD < AB + AD + BC + CD$

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

$AC = AD + DC$

$2(BD) < AB + BC + AC$

$BD < \frac{AB + BC + AC}{2}$

ReplyOptionsExercise 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, $BB' < AB + AB'$

$2(BD) < AB + BC$

$BD < \frac{AB + BC}{2}$

ReplyOptionsExercise 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 $BC < CF + BF$.

$AB + AC + BC < BE + AE + AD + CD + CF + BF$

The sum of the triangle medians is greater than the semi-perimeter.$AE = \frac{AC}{2}$ ; $CD = \frac{BC}{2}$ ; $BF = \frac{AB}{2}$

$AB + AC + BC < BE + \frac{AC}{2} + AD + \frac{BC}{2} + CF + \frac{AB}{2}$

$BE + AD + CF > \frac{AB + AC + BC}{2}$

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