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Full question: Let $\triangle$ ABC be a isosceles triangle (AB = AC) and its area is 501 $cm^2$. BD is the internal bisector of $\widehat{ABC}$ (D $\in$ AC). E is a point on the opposite ray of CA such that CE = CB. I is a point on BC such that CI = 1/2 BI. EI intersects AB at K. BD intersects KC at H. Find $Area_{ \triangle AHC}$.

The figure should look like this: enter image description here Answer: 167 $cm^2$

  • I found this question on a national competition, which ended in May 2016

  • My effort: Let HN $\bot$ BC, HM $\bot$ AB, IL $\bot$ AC, AO $\bot$ BC

The figure should be: enter image description here Then the right triangles $\triangle$ BMH and $\triangle$ BNH are congruent (Hypotenuse-Angle theorem), so HM = HN.

$$Area_{\triangle AHC} = Area_{\triangle ABC} - Area_{\triangle AHB} - Area_{\triangle BHC}$$

$$Area_{\triangle AHC} = 501 - \frac{1}{2}.HM.AB - \frac{1}{2}.HN.BC$$

$$Area_{\triangle AHC} = 501 - \frac{1}{2}.HN.(AB + BC)\\ = 501 - \frac{1}{2}.HN.(AC + CE)\\ = 501 - \frac{1}{2}.HN.AE$$

Here comes the 'crazy' part. A friend told me she could solve HN.AE = 2.IL.AC, then:

$$Area_{\triangle AHC} = 501 - 2.\left(\frac{1}{2}.IL.AC\right)\\ = 501 - 2.Area_{\triangle AIC}$$

$$CI = \frac{1}{2}.BI \implies \frac{CI}{BC} = \frac{CI}{CI + BI} = \frac{\frac{1}{2}.BI}{\frac{1}{2}.BI + BI} = \frac{1}{3}$$

We then have:

$$\frac{Area_{\triangle AIC}}{Area_{\triangle ABC}} = \frac{\frac{1}{2}.AO.IC}{\frac{1}{2}.AO.BC} = \frac{IC}{BC} = \frac{1}{3} \implies Area_{\triangle AIC} = \frac{1}{3}.Area_{\triangle ABC} = \frac{1}{3}.501 = 167 (cm^2)$$

So $Area_{\triangle AHC} = 501 - 2.167 = 167 (cm^2)$

  • 167 $cm^2$ IS the correct answer, but my friend forgot how to solve HN.AE = 2.IL.AC, which I couldn't myself. So can you help me with that, or am I going the wrong direction?

Thank you for checking in. Any help would be appreciated.

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  • $\begingroup$ "EI intersects KC at H"? $\endgroup$ – mathlove Aug 13 '16 at 5:06
  • $\begingroup$ Sorry, fixed that @mathlove $\endgroup$ – William Phoenix Aug 13 '16 at 7:12
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The areas of triangles $AHC$ and $ABC$ are in the ratio $HD/BD$, so I think the easiest way to find the required area is that of computing $HD/BD$.

From $K$ draw a line parallel to $AC$, intersecting $BD$ and $BC$ at $G$ and $F$ respectively (see picture below). Triangles $KFI$ and $ECI$ are similar, which entails $FI={1\over3}KF={1\over3}BK$ and $$ BF=BI-FI={2\over3}BC-{1\over3}BK. $$

By the similarity of $ABC$ and $KBF$ we have $BG=(BF/BC)BD$. Inserting here the above equality yields: $$ BG=\left({2\over3}-{1\over3}{BK\over BC}\right)BD. $$

By the similarity of $KGH$ and $CDH$ we also have $$GH={HK\over HC}HD={BK\over BC}HD,$$ where we have used that ${HK/HC}={BK/BC}$ by the theorem of triangle bisector. In the end we thus get: $$ BD-HD=BH=BG+GH=\left({2\over3}-{1\over3}{BK\over BC}\right)BD+ {BK\over BC}HD, $$ whence $HD={1\over3}BD$.

enter image description here

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  • $\begingroup$ 1. $$\frac{Area_{\triangle AHC}}{Area\{\triangle ABC}} = \left(\frac{HD}{BD}\right)^2, not \frac{BD}{HD}$$. 2, I still don't understand why from the second last line we get HD = 1/3 BD $\endgroup$ – William Phoenix Aug 13 '16 at 15:10
  • $\begingroup$ 1. Triangles $ABC$ and $AHC$ share the same base $AC$, so their areas are in the same ratio as their altitudes, which in turn are in the ratio $BD/HD$. $${}$$ 2. By transporting all terms with $BD$ on the left and all the terms with $HD$ on the right you get $${1\over3}\left(1+{BK\over BC}\right)BD=\left(1+{BK\over BC}\right)HD.$$ $\endgroup$ – Aretino Aug 13 '16 at 15:26
  • $\begingroup$ But BD is the internal bisector, not the altitude? $\endgroup$ – William Phoenix Aug 14 '16 at 1:22
  • $\begingroup$ If you draw the altitudes $BM$ and $HN$ you'll see at once that triangles $DHN$ and $DBM$ are similar, hence $BM/HN=BD/HD$. $\endgroup$ – Aretino Aug 14 '16 at 7:59
  • $\begingroup$ Ok, I understand everything now. Thank you for spending your time to help me and sorry for being quite slow =)) $\endgroup$ – William Phoenix Aug 15 '16 at 5:50

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