INTRODUCING 5 - days-a-week problem solving session for Math Olympiad and ISI Entrance. Learn More 

June 23, 2020

Acute angled Triangle | PRMO II 2019 | Question 29

Try this beautiful problem from the Pre-RMO II, 2019, Question 29, based on Acute angled triangle.

Acute angled triangle - Problem 29


Let ABC be a acute angled triangle with AB=15 and BC=8. Let D be a point on AB such that BD=BC. Consider points E on AC such that \(\angle\)DEB=\(\angle\)BEC. If \(\alpha\) denotes the product of all possible val;ues of AE, find[\(\alpha\)] the integer part of \(\alpha\).

  • is 107
  • is 68
  • is 840
  • cannot be determined from the given information

Key Concepts


Equation

Algebra

Integers

Check the Answer


Answer: is 68.

PRMO II, 2019, Question 29

Higher Algebra by Hall and Knight

Try with Hints


First hint

The pairs \(E_1\),\(E_2\) satisfies condition or \(E_1\)=intersection of CBO with AC and \(E_2\)=intersection of \(\angle\)bisector of B and AC

since that \(\angle DE_2B\)=\(\angle CE_2B\) and for \(E_1\)\(\angle BE_1C\)=\(\angle\)BDC=\(\angle\)BCD=\(\angle BE_1D\)

or, \(AE_1.AC\)=\(AD.AB\)=\(7 \times 15\)

\(\frac{AE_2}{AC}\)=\(\frac{XY}{XC}\)

(for y is midpoint of OC and X is foot of altitude from A to CD)

Acute angled Triangle problem

Second Hint

\(\frac{XD}{DY}=\frac{7}{8}\) and DY=YC

or, \(\frac{XD+DY}{XC}\)=\(\frac{15}{7+8+8}\)=\(\frac{15}{23}\)

or, \(\frac{XY}{XC}=\frac{15}{23}\)

or, \(\frac{AE_2}{AC}\)=\(\frac{15}{23}\)

or, \(AE_1.AE_2\)=\(\frac{15}{23}(7.15)\)=\(\frac{225 \times 7}{23}\)

Final Step

\([\frac{225 \times 7}{23}]\)=68.

Subscribe to Cheenta at Youtube


Leave a Reply

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.
JOIN TRIAL
support@cheenta.com
enter