171. Another "slicing" proposed problems (middle school).

by Virgil Nicula, Nov 17, 2010, 7:14 PM

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=378371

PP1. Let $ABC$ be an $A$-isosceles triangle with $A=100^{\circ}$ . Consider the point $D$ which belongs to the ray $(AB$ so that $AD=BC$ . Ascertain $m\left(\widehat{ADC}\right)$ .

Proof 1 (trigonometric). Denote $m\left(\widehat{ADC}\right)=x$ . Apply the Sinus' theorem in the triangles $ADC$ , $ABC$ . Therefore, $AD=BC\iff$ $\frac {AD}{AC}=\frac {BC}{AC}$ $\iff$

$\frac {\sin (80^{\circ}-x)}{\sin x}=\frac {\sin 100^{\circ}}{\sin 40^{\circ}}$ $\iff$ $\sin (80^{\circ}-x)=2\sin x\cos 40^{\circ}$ $\iff$ $\sin (80^{\circ}-x)=\sin (40^{\circ}+x)-\sin (40^{\circ}-x)$ $\iff$

$\sin (40^{\circ}+x)-\sin (80^{\circ}-x)=\sin (40^{\circ}-x)$ $\iff$ $2\sin (x-20^{\circ})\cos 60^{\circ}=\sin (40^{\circ}-x)$ $\iff$

$\sin (x-20^{\circ})=\sin (40^{\circ}-x)$ $\iff$ $x-20^{\circ}=40^{\circ}-x$ $\iff$ $x=30^{\circ}$ $\iff$ $m\left(\widehat{ADC}\right)=30^{\circ}$ .

Proof 2 (metric). Denote $BC=a\ ,\ AB=AC=b$ and the point $E\in (BC)$ so that $CE=CA$ . Observe that $AD=BC\iff$

$BD=BE=a-b$ . Therefore, $a=2b\cdot\cos 40^{\circ}$ and $DE=EC\iff 2(a-b)\cdot\cos 20^{\circ}=b\iff$ $2\cdot\left(2\cdot\cos 40^{\circ}-1\right)\cdot\cos 20^{\circ}=1$ $\iff$

$2\cdot\left(\cos 60^{\circ}+\cos 20^{\circ}-\cos 20^{\circ}\right)=1$ , O.K. Thus, $DE=EC\iff\triangle DEC$ is $E$-isosceles, i.e. $m(\angle EDC)=10^{\circ}\iff$ $m(\angle ADC)=30^{\circ}$ .

Lemma. Let $ABC$ be a triangle with $B=30^{\circ}$ and $C=20^{\circ}$ . Consider a point $D\in (BC)$ . Prove that $\boxed{BD=AC\ \iff\ DA=DC}$ .

Proof 1 (yetti). Take regular 18-gon $P_1P_2...P_{18}$ with side $p_1$ and diagonals $p_2 < p_3 < ... < p_9$ . From equilateral triangle with base $P_1P_8= p_7,$ $p_1 + p_5 = p_7$ .

Let $X \equiv P_1P_6 \cap P_4P_{11}$ $\Longrightarrow$ $\triangle XP_1P_4$ is isosceles with base $XP_4$ and base angles $80^\circ$ $\Longrightarrow$ $\triangle XP_6P_{11}$ is isosceles with $XP_{11} = P_6P_{11} = p_5$ $\Longrightarrow$

$XP_4 = p_1 = P_3P_4$ $\Longrightarrow$ $\triangle XP_3P_4$ is isosceles with base $XP_3$ and base angles $40^\circ$ $\Longrightarrow$ $X \equiv P_4P_{11} \cap P_3P_{8}\implies X\in P_1P_6\cap P_3P_8\cap P_4P_{11}$ . By

symmetry $XP_3 = XP_6$ . Put $B \equiv P_1, C \equiv P_6, A \equiv P_3.$ Then $DA \equiv DP_3 = DP_6 \equiv DC$ $\Longleftrightarrow$ $D \equiv X$ $\Longleftrightarrow$ $BD \equiv P_1D = P_1P_4 = P_3P_6 \equiv AC$ .

Proof 2 (trigonometric). Denote $m(\angle DAC)=x$ . Thus, $\boxed{BD=AC}\iff$ $\frac {BD}{DA}=\frac {AC}{DA}$ $\iff$ $\frac {\sin (50^{\circ}+x)}{\sin 30^{\circ}}=$ $\frac {\sin (20^{\circ}+x)}{\sin 20^{\circ}}$ $\iff$

$\sin (20^{\circ}+x)=$ $2\sin 20^{\circ}\sin (50^{\circ}+x)$ $\iff$ $\sin (20^{\circ}+x)=$ $\cos (30^{\circ}+x)-\cos (70^{\circ}+x)$ $\iff$ $\sin (60^{\circ}-x)-$

$\sin (20^{\circ}+x)=$ $\sin (20^{\circ}-x)$ $\iff$ $2\sin (20^{\circ}-x)\cos 40^{\circ}=$ $\sin (20^{\circ}-x)$ $\iff$ $x=20^{\circ}$ $\iff$ $\boxed{DA=DC}$ .

Proof 3. Construct the circumcircle $(O)$ of the $\triangle ABC$ . Suppose $\boxed{BD=AC}$ . Therefore, $m(\angle AOB)=40^\circ\Longrightarrow m(\angle OBA)=70^\circ\Longrightarrow$

$m(\angle OBD)=40^\circ$ $\stackrel{OB=BD}{\Longrightarrow}m(\angle BOD)=70^\circ\Longrightarrow m(\angle AOD)=30^\circ\Longrightarrow$ $\triangle AOD\stackrel{(LUL)}{\equiv}\triangle COD\Longrightarrow \boxed{AD=CD}$ .

Suppose $\boxed{DA=DC}$ . Therefore obtain that $\triangle AOD\stackrel{(LLL)}{\equiv}\triangle COD\Longrightarrow$ $m(\angle COD)=m(\angle DBA)=30^{\circ}$ and

$m(\angle OCD)=m(\angle BDA)=40^{\circ}$ . From $AD=DC$ $\stackrel{(ULU)}{\Longrightarrow}\triangle ABD\equiv\triangle DOC\Longrightarrow BD=OC=AC\implies \boxed {BD=AC}$ .

Proof 3 (synthetic). Denote $BC=a\ ,\ AB=AC=b$ and the point $E\in (BC)$ so that $CE=CA$ . Observe that $AD=BC\iff$

$BD=BE=a-b$ and $m(\angle DAE)=30^{\circ}\ ,\ m(\angle ADE)=20^{\circ}$ . Apply the above lemma to $\triangle EAD$ and the point $B\in (AD)$

for which $BD=BE$ and obtain that $DE=AB$ , i.e. $DE=EC$ $\implies$ $m(\angle EDC)=m(\angle ECD)=10^{\circ}$ , i.e. $m(\angle ADC)=30^{\circ}$ .


PP2 (Nicolae Coculescu's Contest juniors 2010). Let $ABC$ be a triangle with $B=60^{\circ}$ and $C=40^{\circ}$ .

Consider the point $D\in (BC$ for which $CD=2s$ , where $2s=a+b+c$ . Ascertain $m(\angle ADB)$ .

Proof 1 (synthetic). Denote $E\in (CD)$ so that $CE=b$ . Then $m(\angle CAE)=m(\angle CEA)=20^{\circ}$ . Denote $G\in(AE)$ so that $AG=c$ . Thus, $m(\angle ABG)=$

$m(\angle AGB)=40^{\circ}$ . Observe that the quadrilateral $AGCB$ is cyclically, i.e. $m(\angle BGC)=m(\angle BAC)=m(\angle GCB)=80^{\circ}$ $\implies$ $GB=a$ $\implies$

$\triangle GBE$ is isosceles $\implies$ $GE=GB=a$ . Therefore, $AE=a+c=ED$ , i.e. $\triangle AED$ is isosceles $\implies$ $m(\angle EDA)=m(\angle EAD)=10^{\circ}$ .
This post has been edited 47 times. Last edited by Virgil Nicula, Dec 1, 2015, 11:18 AM

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26. Outside of a triangle.
25. Geometric equations.
24. A fine and cool inequalities.
23. A very nice exponential inequality (own - from CRUX).
22. Some interesting limits (sequence/function).
21 bis. Some problems with complex numbers.
21. Probleme de geometrie (Yetty & I).
20. O ineg.cu suma de AG/AI .
19. Own proposed problems in CRUX Mathematicorum - Canada.
18. O problema a lui Toshio Seimiya.
17. O inegalitate "tare" intr-un triunghi.
16. Length of the (interior) angle bisector (10 proofs).
15. Inequalities stronger than Panaitopol's.
14. Opinii si comentarii relativ la inv. rom.
13. Inegalitatea de la Sorin Borodi.
12. Inegalitatea I. Maftei & S. Radulescu (2).
11. Inegalitatea I. Maftei & S. Radulescu (1).
10. Walker's inequality and its extension.
9. Inegalitatea HO+HA+HB+HC >= 3R .
8. The first problem Shortlist ONM 2010 Romania.
7. Some interesting "slicing" problems.
6. Interesting problems from JBMO.
5. Theoretical preliminary for inequalities.
4. Remarkable geometrical inequalities (2).
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    by ryanbear, May 9, 2023, 6:10 AM

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    by OlympusHero, Sep 14, 2022, 4:44 AM

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    by tigerzhang, Aug 1, 2021, 12:02 AM

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    by GoogleNebula, Apr 14, 2021, 5:25 AM

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    by samrocksnature, Apr 14, 2021, 5:16 AM

  • Holy- Darn this is good. shame it's inactive now

    by the_mathmagician, Jan 17, 2021, 7:43 PM

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    by OlympusHero, Dec 30, 2020, 6:08 PM

  • long blog

    by MrMustache, Nov 11, 2020, 4:52 PM

  • 372554 views!

    by mrmath0720, Sep 28, 2020, 1:11 AM

  • wow... i am lost.

    369302 views!

    -piphi

    by piphi, Jun 10, 2020, 11:44 PM

  • That was a lot! But, really good solutions and format! Nice blog!!!! :)

    by CSPAL, May 27, 2020, 4:17 PM

  • impressive :D
    awesome. 358,000 visits?????

    by OlympusHero, May 14, 2020, 8:43 PM

72 shouts
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  • Joined: Jun 22, 2005
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  • Blog created: Apr 20, 2010
  • Total entries: 456
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