Problem:
Point P is inside â–³ABC. Line segments APD,BPE and CPF are drawn with D on BC,E on CA, and F on AB (see the figure at the right). Given that AP=6,BP=9,PD=6,PE=3 and CF=20, find the area of â–³ABC.
Solution:
We are given length information about the three segments through P. Our strategy is to translate one of these segments to form a new triangle (inside of â–³ABC ) for which we know all three sides, and hence the area. We then multiply this area by an appropriate ratio to obtain the area of â–³ABC.
We first find the lengths of CP and PF. To this end, observe that
Area(△BAC)Area(△BPC)​=ADPD​=6+66​=21​(1)
and
Area(△ABC)Area(△APC)​=BEPE​=3+93​=41​(2)
Thus
CFPF​​=Area(△ACB)Area(△APB)​=Area(△ACB)Area(△ACB)−Area(△APC)−Area(△BPC)​=1−41​−21​=41​​(3)
Since CF=20, it follows that PF=5 and CP=15. Furthermore,
CDBD​=Area(△ACD)Area(△ADB)​​=Area(△PDC)Area(△PDB)​=Area(△ACD)−Area(△PDC)Area(△ADB)−Area(△PDB)​=Area(△ACP)Area(△APB)​=1,​
where the last equality results from dividing equation (3) by equation (2).
Next construct DG parallel to CF, with G on PB as shown. We show that △GDP is a right triangle with sides of 29​,6 and 215​ and then show that the area of △GDP is 81​ of the area of △ABC. It will then follow that the desired answer is 8⋅227​=108.
To establish the above claims, first note that △BDG∼△BCP. Since BD=21​BC, the sides of these two triangles are in a ratio of 1:2. It follows that DG=21​PC=215​ and PG=GB=21​PB=29​. Since PD=6 was given, we see that △GDP is a right triangle as claimed. Next note that
Area(△GBD)Area(△PBC)​=(12​)2=4 and Area(△GPD)Area(△GBD)​=PGBG​=1
Using these ratios with (1) gives
Area(△ABC)​=Area(△PBC)Area(△ABC)​⋅Area(△GBD)Area(△PBC)​⋅Area(△GPD)Area(△GBD)​⋅Area(△GPD)=2⋅4⋅1⋅227​=108​
Note. Implicit in this solution is a method for constructing â–³ABC with straightedge and compass from the given data. The reader is invited to explore the conditions such data must satisfy in order to ensure that â–³ABC exists (and is unique).
OR
Let P=(0,0),D=(6,0),A=(−6,0),E=(h,k) and B=(−3h,−3k). Solving the equations for AE and BD simultaneously, we find C=(3h+12,3k). Next, the coordinates of F can be found by solving the equations of CP and AB simultaneously; the result is F=(−4−h,−k). Finally, solving the equations h2+k2=9 and (4+h)2+k2=25 (arising from PE=3 and CF=20, respectively) one finds that h=0 and k=3. Once we have the coordinates of A,B and C, we can find that the area of △ABC is 108​.
The problems on this page are the property of the MAA's American Mathematics Competitions
