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

But try the problem first…

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)

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.

## Other useful links

- https://www.cheenta.com/rational-number-and-integer-prmo-2019-question-9/
- https://www.youtube.com/watch?v=lBPFR9xequA

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