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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




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


(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.

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