Let A B C be a triangle let D be a point on the segment B C such that A D=B C . Suppose C A D=x^ , A B C=y^ and A C B=z^ and x, y, z are in an arithmetic progression in that order where the first term and the common

Correct answer: 59 (A D)/( y)=(B D)/( (x+y+z)) (A D)/( z)=(C D)/( x) aligned & ( (x+y+z))/( y)+( x)/( z)=1 & ( (3 y))/( y)+( (y-d))/( (y+d))=1 & ( (y-d))/( (y+d))=-2 2 y & (3 y+d)=0 3 y+d=180^ , 360^ & y & d are integers d is multiple of 3 & If 3 y+d=180^ y_ =59^ & 3 y+d=360^ y_ =119^ (if y is obtuse, then z must be acute) & Only 1^ st case is possible y_ =59^ aligned

Olympiad workbook

Exam: Olympiad workbook
Subject: IOQM
Chapter: Properties of Triangles

Olympiad workbook · IOQM · Properties of Triangles

Let A B C be a triangle let D be a point on the segment B C such...

Let

A B C

be a triangle let

D

be a point on the segment

B C

such that

A D=B C

. Suppose

\angle C A D=x^{\circ}, \angle A B C=y^{\circ}

and

\angle A C B=z^{\circ}

and

x, y, z

are in an arithmetic progression in that order where the first term and the common difference are positive integers. Find the largest possible value of

\angle A B C

in degrees. Let A B C be a triangle let D be a point on the segment B C such that A D=B C . Suppose C A D=x^ , A B C=y^ and A C

Correct answer

Step-by-step solution

\frac{A D}{\sin y}=\frac{B D}{\sin (x+y+z)}

\frac{A D}{\sin z}=\frac{C D}{\sin x}

\begin{aligned} & \frac{\sin (x+y+z)}{\sin y}+\frac{\sin x}{\sin z}=1 \\ & \frac{\sin (3 y)}{\sin y}+\frac{\sin (y-d)}{\sin (y+d)}=1 \\ & \frac{\sin (y-d)}{\sin (y+d)}=-2 \cos 2 y \\ & \sin (3 y+d)=0 \Rightarrow 3 y+d=180^{\circ}, 360^{\circ} \\ & y & d \text { are integers } \Rightarrow d \text { is multiple of } 3 \\ & \text { If } 3 y+d=180^{\circ} \Rightarrow y_{\max }=59^{\circ} \\ & 3 y+d=360^{\circ} \Rightarrow y_{\max }=119^{\circ} \text { (if } y \text { is obtuse, then } z \text { must be acute) } \\ & \text { Only } 1^{\text {st }} \text { case is possible } y_{\max }=59^{\circ}\end{aligned}

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