Incenter Beast

Geometry Level 5

In A B C \triangle ABC , I I is the incenter of the triangle. A C + A I = B C AC+AI=BC and A B + B I = A C AB+BI=AC . Find the measure of A B C \angle ABC , to the nearest degree.


The answer is 51.

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2 solutions

Lemuel Liverosk
May 11, 2016

A more precise answer should be 360 7 ° 51.43 ° \frac{360}{7}°\approx 51.43° .

Cut B C \overline{BC} into B R \overline{BR} and C R \overline{CR} , where C R = C A CR=CA . As A C + A I = B C AC+AI=BC , A I = B R AI=BR .

A C R \triangle ACR is isosceles, with C I \overline{CI} as an angle bisector. Therefore, C I \overline{CI} is also the perpendicular bisector of A R \overline{AR} , meaning that A I = R I AI=RI , m C R I = m C A I m\angle CRI=m\angle CAI .

m C R I = m R B I + m R I B m\angle CRI=m\angle RBI+m\angle RIB . As R I = I A = R B RI=IA=RB , m R B I = m R I B m\angle RBI=m\angle RIB , then m C R I = 2 m R B I = m R B A m\angle CRI=2m\angle RBI=m\angle RBA .

On the other hand, m C A I = m B A I m\angle CAI=m\angle BAI , then m R B A = m I A B m\angle RBA=m\angle IAB . A B I B A R \triangle ABI\cong \triangle BAR ( S . A . S . S.A.S. ). Thus m C A B = 2 m A B C m\angle CAB=2m\angle ABC .

By going through the same process for a second time, we can get B I = I S = C S B I S C S I m A B C = 2 m A C B BI=IS=CS→\triangle BIS\cong \triangle CSI→m\angle ABC=2m\angle ACB .

Therefore, m C A B : m A B C : m B C A = 4 : 2 : 1 m\angle CAB:m\angle ABC:m\angle BCA=4:2:1 , making m A B C = 2 × 180 ° 7 = 360 7 ° m\angle ABC=2\times \frac{180°}{7}=\frac{360}{7}° .

Nice problem.Took me quite a long time to solve.Please post more such challenging geometry problems.

Indraneel Mukhopadhyaya - 5 years, 1 month ago

Awesome problem! Keep posting such problems :)

A Former Brilliant Member - 5 years, 1 month ago
Choi Chakfung
May 13, 2016

P r o d u c e C B t o E w h i c h D B = B E . L e t A C B = θ , C A E = C E A = 90 ° θ 2 D i s i n c e n t r e . D A B + D B A = 90 ° θ 2 A D B = 90 ° + θ 2 . A D B + A E B = 180 ° A , D , B , E i s c o n c y c l i c . D B = B E D A B = B A E = D A C l e t D A B = α C A E = C E A = 3 α . A C E = 180 ° 6 α A C D = D C B = 90 ° 3 α A B C = B A E + A E B = 4 α P r o d u c e B A t o F w h e r e A F = D A B F = B C B F C = B C F = 90 2 α F C D = F C B D C B = ( 90 ° 2 α ) ( 90 3 α ) = α F C A = 90 ° 2 α ( 180 ° 6 α ) = 4 α 90 D C F = D A B = α . F C D A i s c o n c y c l i c A F = A D F C A = A C D 4 α 90 ° = 90 ° 3 α α = 25.714... B A B = 2 α = 51.428... Produce\quad CB\quad to\quad E\quad which\quad DB=BE.Let\quad \angle ACB=\theta ,\angle CAE=\angle CEA=90°-\frac { \theta }{ 2 } \\ \because D\quad is\quad incentre.\angle DAB+\angle DBA=90°-\frac { \theta }{ 2 } \angle ADB=90°+\frac { \theta }{ 2 } .\angle ADB+\angle AEB=180°\\ \therefore A,D,B,E\quad is\quad concyclic.DB=BE\Rightarrow DAB=BAE=DAC\\ let\quad DAB=\alpha \quad \Rightarrow CAE=CEA=3\alpha .\angle ACE=180°-6\alpha \Rightarrow ACD=DCB=90°-3\alpha \\ ABC=BAE+AEB=4\alpha \\ Produce\quad BA\quad to\quad F\quad where\quad AF=DA\Rightarrow BF=BC\Rightarrow BFC=BCF=90-2\alpha \\ \angle FCD=FCB-DCB=(90°-2\alpha )-(90-3\alpha )=\alpha \\ FCA=90°-2\alpha -(180°-6\alpha )=4\alpha -90\\ \angle DCF=\angle DAB=\alpha .\because FCDA\quad is\quad concyclic\\ AF=AD\Rightarrow \angle FCA=\angle ACD\Rightarrow 4\alpha -90°=90°-3\alpha \\ \Rightarrow \alpha =25.714...\angle BAB=2\alpha =51.428...

Nice solution

Ayush Kumar - 2 years, 9 months ago

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